text:
The temporal indicator of 1648-1715 represents a chronological context associated with resources found in Massey University Library, University of New Zealand.
The Resource 1648-1715
100+ Items that are about the Temporal 1648-1715
A Dialogue between the confederate princes concerning the present affairs of Europe, (electronic resource)
A Letter from a true Dutchman to his friend : exhibiting the reasons why the confederates ought to refuse the peace, which is now offered to them by France, (electronic resource)
A Letter to a friend concerning the partition treaty, (electronic resource)
A Letter to the Paris Gazetteer upon the siege and taking of Namur, by the author of the Safety of Europe, (electronic resource)
A Relation of the battle of Landen, (electronic resource)
A Second letter to a friend concerning the partition treaty : with his answer, (electronic resource)
A brief description of the future history of Europe, from Anno 1650 to An. 1710. : Treating principally of those grand and famous mutations yet expected in the world, as, the ruine of the Popish hierarchy, the final annihilation of the Turkish Empire, the conversion of the eastern and western Jews, and their restauration to their ancient inheritance in the Holy Land, and the Fifth Monarchie of the universall reign of the Gospel of Christ upon Earth. With principal passages upon every of these, out of that famous manuscript of Paul Grebner, extant in Trinity-Colledge Library in Cambridge. Composed upon the occasion of the young Kings arrival into Scotland, to shew what will in probability be the event of the present affairs in England and Scotland, (electronic resource)
A brief description of the future history of Europe, from anno 1650 to an. 1710 : treating principally of those grand and famous mutations yet expected in the world, as the ruine of the Popish hierarchy, the final annihilation of the Turkish Empire, the conversion of the eastern and western Jews and their restauration to their ancient inheritance in the Holy Land, and the fifth monarchie of the universall reign of the Gospel of Christ upon earth : with principal passages upon every of these, out of that famous manuscript of Paul Grebner, extant in Trinity-College Library in Cambridge : composed upon the occasion of the young king's arrival into Scotland, to shew what will in probability be the event of the present affairs in England and Scotland, (electronic resource)
A briefe relation of some affaires and transactions, civill and military, both forraigne and domestique, (electronic resource)
A compleat history of Europe, or, A view of the affairs thereof, civil and military from the beginning of the Treaty of Nimeguen, 1676, to the conclusion of the peace with the Turks, 1699 : including the articles of the former, and the several infringements of them, the Turkish Wars, the forming of the Grand Confederacy, the revolution in England, &c. : with a particular account of all the actions by sea and land on both sides, and the secret steps that have been made towards a peace, both before, as well as during the last negotiation : wherein are the several treaties at large, the whole intermix'd with divers original letters, declarations, papers and memoirs, never before published, written by a gentleman, who kept an exact journal of all transactions, for above these thirty years, (electronic resource)
A compleat history of Europe, or, A view of the affairs thereof, civil and military, from the beginning of the Treaty of Nimeguen, 1676, to the conclusion of the Peace at Reswick, 1697 : including the articles of the former and the several infringments of them, the Turkish wars, the forming of the Grand Confederacy, the revolution in England, &c. ... the whole intermix'd with divers original letters, papers, and memoirs never before published, written by a gentleman who kept an exact journal of all transactions for above these twenty years, (electronic resource)
A declaration of His Electoral Highness the Duke of Brandenburgh, concerning the present war with France: : directing his subjects and vassals, and all other his dependants, how they are to demean themselves in their trading, negotiation, and all incident occasions during the present war. As also a letter sent from the Imperial Diet at Ratisbon to the thirteen cantons in Switzerland, March 7, 1689. Exhorting them to adhere to His Imperial Majesty, and the princes of the empire, in the present war against France, (electronic resource)
A declaration of the proceedings of thirteen Christian kings, princes, and dukes, and the present state and condition of the King of Scots: : with their summons to all Christendom, for the raising of three hundred thousand horse and foot, under the command of Prince Philip, the third son to the Queen of Bohemia. And the King of Scots message to the Emperor of Germany, and the princes of Italy; sent by the Lord Wilmot: with their royal answer thereunto, (electronic resource)
A dialogue between the French King, and the late King James : at St. Germains en Laye: occasion'd by the signing of the peace, (electronic resource)
A dialogue between the King of France and the late King James, occasion'd by the death of the Queen. Written originally in French, at Paris, (electronic resource)
A dialogue between the late King James and the Prince of Conty, (electronic resource)
A geographical and historical description of those parts of Europe which are the seat of war : Viz. Flanders, the Rhine, Savoy, Catalonia, Hungary, Greece. With an account of the several revolutions of those places, either by conquest or treaties: particularly, from the Treaty of Munster, 1648. to this present time. By A.B. Gent. With maps of those countries, and the parts adjacent, (electronic resource)
A history of Europe from 1610 to 1715, by W.F. Reddaway
A letter from Monsieur de Cros (who was an embassador at the Treaty of Nimeguen and a resident in England in K. Ch. the Second's reign) : which may serve for an answer to the impostures of Sir. Wm. Temple, heretofore ambassador from England at the Hague and at Nimeguen ... : together with some remarks upon his memoirs, to make appear how grosly he is mistaken in the greatest part of the most important matters he relates concerning what passed from the year 1672 until the year 1679, (electronic resource)
A letter from Monsieur de Cros : (who was an embassador at the Treaty of Nimeguen and a resident at England in K. Charles the Second's reign) to the Lord ----- : being an answer to Sir Wm. Temple's memoirs concerning what passed from the year 1672 until the year 1679, (electronic resource)
A letter from the King of Spain written to the Pope of Rome : Licens'd, September 29. 1690, (electronic resource)
A letter written by King James to his Holiness the Pope, : as 'tis extracted out of the registers taken at Dublin, (electronic resource)
A modern view of such parts of Europe : that hath lately been and still are the places of great transactions, viz. Italy with all its principalities. France with all its provinces and bishopricks. Germany with the Dukedome of Lorrain, and all the electorates, and lordshops of the empire. Spain, with all its dominions, &c. Wherein is shewed the present state of all those countries, with curious remarks of antiquity interwoven, (electronic resource)
A new survey of the present state of Europe : containing remarks upon several soveraign and republican states : with memoires historical, chronological, topographical, hydrographical, political, &c, by Gidion Pontier, &c. ; done into English by J.B. Doctor of Physick, (electronic resource)
A panegyrick mercury upon the successes against the Turks : in the Autumn, 1683, (electronic resource)
A particular advice from foreign parts : also from the office of intelligence ., (electronic resource)
A perfect diurnall, or Occurrences of certain military affairs in order to the armies in England, Scotland, and Ireland. : With variety of the transactions of other nations, (electronic resource)
A relation of the battel [sic] of Landen : Being a collection of the several accounts thereof. Published by authority, (electronic resource)
A relation of the battel [sic] of Landen. Published by authority., (electronic resource)
A representation of the present affairs and interests of the most considerable parts of Europe, more especially of those of the Netherlands : as they now stand, in the beginning of the year 1677. Laid open in a letter from Holland. By a lover of truth and peace, (electronic resource)
A review of what God hath been pleased to do this year, according to prophecy : and particularly near the 23d of the month we stile August : in the general peace, and the Muscovian and imperial victories over the Turks : with a most humble apology and defence concerning the falling short of events, of what was expected according to the interpretation of that sure word of prophecy, by T. Beverley, (electronic resource)
A treatise of the interest of the princes and states of Christendome, written in French by the the most noble and illustrious Prince, the Duke of Rohan ; Englished by H.H, (electronic resource)
A true and exact relation of the great victory obtained by General Schults over Count Teckely in the Upper Hungaria, on the 20th of Septemb. 1684. : Together with an account of the battle between the Duke of Lorraine and the Serasquier Bassa before Buda, (electronic resource)
A true history of all the memorable transactions that have happen'd in England, Scotland, Ireland, Flanders, &c. relating to state affairs, sieges, battles, sea-fights, and other naval matters : With a particular relation of all the plots and conspiracies against the life of King William and his government, at home and abroad: also by what means they were discovered and defeated, by signal providences: from his coming to the crown, to this present year, 1696. Especially a large account of the late horrid conspiracy, to assassine his sacred life, and bring in a French power. Likewise the confessions and executions of divers of the conspirators, and other things worthy of note, (electronic resource)
An Account of last Thursday's sea engagement between the Dutch and French fleets near the coast of Brest : with the total routing of the French fleet, (electronic resource)
An Account of the actions of the confederate and French armies. : With a list of the German forces arrived to the assistance of the Duke of Savoy., In a letter from a gentleman in the army to his friend in London, (electronic resource)
An Account of the late actions at sea, between the Saint Alban's and the Happy Return, two of Their Majesties men of war, and twenty four sail of French ships : with their convoy of 36 guns, near the port of Cherbourg : with an account of the sinking the convoy, [sic] and taking twenty sail of merchant men : in a letter from a merchant at Plymouth, (electronic resource)
An Account of the late engagement at sea, on the coast of Spain, between two of Their Majesties' fifth rate frigates with two French men of war and a privateer : as also several rich merchant-men, the stranding one [sic] of the French men of war, and taking three other ships richly laden with great numbers of prisoners : in a letter from a merchant at Falmouth to his frind [sic] in London, (electronic resource)
An Answer to a scurrilous pamphlet, lately printed, intituled, A letter from Monsieur de Cros, to the Lord ----, (electronic resource)
An exact accompt of the daily proceedings in Parliament. : With occurrences from foreign parts, (electronic resource)
An exact accompt of the publick transactions of the three nations. : With occurrences from foreign parts ., (electronic resource)
An historical account of the memorable actions of the most glorious monarch William III, King of England, Scotland, France and Ireland, Prince of Orange &c. : giving a true relation of all that hapned of consequence since his glorious expedition into England ... together with the names of several of those worthy persons upon whom he has conferred honours and places of trust since his being proclaimed, (electronic resource)
An historical account of the memorable actions of the most glorious monarch William III, King of England, Scotland, France and Ireland, Prince of Orange &c. : giving a true relation of all that happened of consequence since his glorious expedition into England ... together with the names of several of those worthy persons upon whom he has conferred honours and places of trust since his being proclaimed, (electronic resource)
Annals of the universe; : containing an account of the most memorable actions, affairs, and occurrences which have happen'd in the world: but especially in Europe. From the year 1660. where Mr. Whitlock leaves off, to the year 1680. In two decades: with an index to the whole. Being a continuation of the said Mr. Whitlock's Memorials, (electronic resource)
Articles of peace between the emperour and empire, and France : concluded in the royal palace of Reswick, in Holland, October 30, 1697, (electronic resource)
Bloudy newes from Holland: : being a true relation of the present proceedings of the Dutch-men against the English, at the Hague, Amsterdam, Rotterdam, Brill, Dort, Flushin, and divers other places. Also, a great fight at sea, between the Parliaments ships, and the Hollanders; the number burnt, sunk, and taken, by Captain Johnson and the rest of the English; with the manner of the said engagement. Likewise Lieu. Colonel John Lilburn made Captain of a man of War for the States of Holland, and his letter to the Lord General Cromwel, (electronic resource)
Certain passages of every dayes intelligence from the Army, and his Highness the Lord Protector, and his Council : with other remarkable proceedings in England, Holland, France, Scotland, and Ireland, (electronic resource)
Discourses on the present state of the Protestant princes of Europe : exhorting them to an union and league amongst themselves against all opposite interest, from the great endeavours of the court of France and Rome to influence all Roman Catholick princes, against the Protestant states and religion, and the advantage that our divisions give to their party : wherein the general scope of this horrid Popish Plot is laid down, and presented to publick view, by Edmund Everard ., (electronic resource)
Discourses on the present state of the Protestant princes of Europe: : exhorting them to an union and league amongst themselves, against all opposite interest, from the great endeavours of the court of France and Rome to influence all Roman Catholick princes, against the Protestant states and religion; and the advantage that our divisions give to their party; wherein the general scope of this horrid Popish Plot is laid down, and presented to publick view., By Edmund Everard Esquire. Kept four years close prisoner in the Tower by the contrivance of some English subjects plotting against us in France, whom he five years since discovered; and was lately justified and released by his Majesty, (electronic resource)
Discourses upon the modern affairs of Europe : tending to prove that the illustrious French monarchy may be reduced to terms of greater moderation, (electronic resource)
Dynastic identity in early modern Europe : rulers, aristocrats and the formation of identities, edited by Liesbeth Geevers, Mirella Marini
Early modern Europe from about 1450 to about 1720
Europe a slave unless England break her chains : discovering the grand designs of the French in Flanders, Germany and England for several years past, (electronic resource)
Europe a slave, unless England break her chains : discovering the grand designs of the French-Popish party in England for several years past, (electronic resource)
Europe a slave, unless England break her chains. : Discovering the grand designs of the French-popish-party in England for several years past, (electronic resource)
Europe in the age of Louis XIV
Europe unfolding, 1648-1688, [by] John Stoye
Europe's chains broke, or, A sure and speedy project to rescue her from the present usurpations of the tyrant of France, (electronic resource)
Europe's glory, or, Peace & plenty to the people thereof : being a projection or a scheam [sic] of reasonable terms for establishing a firm and general peace between the monarch of Great Britain, including all the confederate princes, and the French king ..., by Thomas Houghton, of Lyme-street, gent, (electronic resource)
Europäische Geschichtskulturen um 1700 : zwischen Gelehrsamkeit, Politik und Konfession, herausgegeben von Thomas Wallnig [and others], (electronic resource)
Europæ modernæ speculum, or, A view of the empires, kingdoms, principalities, seignieuries [sic], and common-wealths of Europe : in their present state, their government, policy, different interest and mutual aspect one towards another, from the treaty at Munster, anno 1648, to this present year, (electronic resource)
Europæ modernæ speculum, or, A view of the empires, kingdoms, principalities, seignieuries, and common-wealths of Europe : in their present state, their government, policy, different interest, and mutual aspect one towards another, from the treaty at Munster, Anno 1648 to this present 1665, (electronic resource)
Foreign advices, (electronic resource)
Gender, power and privilege in early modern Europe, Jessica Munns and Penny Richards
Good news from the English fleet : being an account of a great bloody engagement which hapned yesterday betwixt Their Majesties fleet commanded by the Earl of Torrington and the French fleet, near the Beachy, upon the coast of Suffolk : with a particular relation of the blowing-up of six of the biggest of the French men of war ., (electronic resource)
Good news from the English fleet: : being an account of a great and bloody engagement which happened yesterday, betwixt Their Majesties fleet, commanded by Admiral Killigrew, and the French fleet, near the Beachy: with a particular account of the taking six of their ships, and sinking three. September 17. Printed according to order, (electronic resource)
Great news from His Majesty's camp near Louvain in Flanders : being a full and true account of the taking a party of the French king's Horse, among whom was the French king's general the Duke of Luxemburgh, in a letter from a gentleman in His Majesty's camp to his friend at London, (electronic resource)
Habsburg and Bourbon Europe, 1470-1720, Roger Lockyer
His Most Catholick Majesty's answer to the Pope's proposals for a general peace throughout Europe, (electronic resource)
Letters written by Sir W. Temple, Bart., and other ministers of state, both at home and abroad : containing an account of the most important transactions that pass'd in Christendom from 1665-1672 : in two volumes, review'd by Sir W. Temple sometime before his death ; and published by Jonathan Swift ., (electronic resource)
Letters written by a French gentleman, giving a faithful and particular account of the transactions at the court of France, relating to the publick interest of Europe : with historical and political reflexions on the ancient and present state of that kingdom, communicated by Monsieur Vassor, (electronic resource)
London Septemb. 27. : Yesterday morning arrived three Holland mails, which bring the following advices. Vienna, September 18, (electronic resource)
Louis XIV en son temps, 1661-1715
Luther's legacy : the Thirty Years War and the modern notion of 'state' in the empire, 1530s to 1790s, Robert von Friedeburg
Memoirs of what past in Christendom : from the war begun 1672 to the peace concluded 1679. Licensed, Novemb. 14. 1691, (electronic resource)
Memoirs of what past in Christendom : from the war begun 1672, to the peace concluded 1679, (electronic resource)
Memoirs of what past in Christendom : from the war begun 1672. To the peace concluded 1679, (electronic resource)
Memoirs of what past in Christendom, from the war begun 1672 to the peace concluded 1679, (electronic resource)
Memoirs of what past in Christendom, from the war begun 1672 to the peace concluded 1679, by Sir William Temple baronet : beginning exactly where his letters leave off, (electronic resource)
Mercurius Britannicus : Communicating his most impartial intelligence from all parts, touching all humours, affairs, and designes, in the three nations, more especially from France, Holland, and Denmark: with the resolves and debates of Parliament, (electronic resource)
Mercurius democritus : communicating faithfully the affairs both in city and countrey, with most observable occurrences from forreign parts, (electronic resource)
Mercurius pragmaticus : Taking a serious view of the present condition, proceedings, and conspiracies of the distracted part of the world; impartially communicating publique affairs, and gently correcting domestick errors, (electronic resource)
Mercurius pragmaticus, : Communicating his most impartial intelligence from all parts, more especially touching the real state, and general affairs of England, Scotland, Ireland, France, and Holland, (electronic resource)
Mercurius pragmaticus. : Communicating his most impartial intelligence from all parts : more especially touching the real state, and general affairs of England, Scotland, Ireland, France, and Holland, (electronic resource)
Mercurius publicus ..., (electronic resource)
Mercurius reformatus: or, The new observator., (electronic resource)
Mercurius republicus. : Communicating martiall actions, with other publique affairs to the nations, of England and Ireland, (electronic resource)
Mercurius veridicus. : Communicating some choice inteligence domestick and forreign, (electronic resource)
Miscellanea. : I. A survey of the constitutions and interests of the Empire, Sueden, Denmark, Spain, Holland, France, and Flanders; with their relation to England, in the year 1671. II. An essay upon the original and nature of government. III. An essay upon the advancement of trade in Ireland. IV. Upon the conjecture of affairs in Octob. 1673. V. Upon the excesses of grief. VI. An essay upon the cure of the gout by Moxa., By a person of honour, (electronic resource)
Moderate occurrences : Faithfully communicating the impartial proceedings of the English and Dutch navies at sea, the affaires and designes now on foot in England, Scotland, France, Ireland, Denmark, Holland, and other parts in Europe, (electronic resource)
Nero gallicanus, or, The true pourtraicture of Lewis XIV : wherein the present war with France is justified, from the necessity of reducing that most Christian king to a more Christian temper, (electronic resource)
New worlds? : transformations in the culture of international relations around the Peace of Utrecht, edited by Inken Schmidt-Voges and Ana Crespo Solana
Occurrences from forraigne parts ..., (electronic resource)
Perfect and impartial intelligence, : of the affairs, in England, Scotland, and Ireland, and other parts beyond the seas, (electronic resource)
Perfect diurnall occurrences of certain military affairs in order to the armies, in England, Scotland, and Ireland. : With variety of the transactions of other nations, (electronic resource)
Perfect occurrences, faithfully communicating ..., (electronic resource)
Perfect proceedings of state-affaires : in England, Scotland and Ireland, with the transactions of other nations, (electronic resource)
Prince Eugene : twin marshall with Marlborough
Context of 1648-1715
Europe -- Foreign relations -- 1648-1715
Europe -- History -- 1648-1715
Europe -- History -- 1648-1715 -- Bibliography
Europe -- History -- 1648-1715 -- Early works to 1800
Europe -- History -- 1648-1715 -- Historiography
Europe -- History -- 1648-1715 -- Newspapers | Early works to 1800
Europe -- History -- 1648-1715 -- Periodicals
Europe -- History -- 1648-1715 -- Prophecies -- Early works to 1800
Europe -- History -- 1648-1715 -- Sources
Europe -- Politics and givernment -- 1648-1715 -- Poetry | Early works to 1800
Europe -- Politics and government -- 1648-1715
Europe -- Politics and government -- 1648-1715 -- Early works to 1800
Europe -- Politics and government -- 1648-1715 -- Humor -- Early works to 1800
Europe -- Politics and government -- 1648-1715 -- Humor | Early works to 1800
Europe -- Politique et gouvernement -- 1648-1715
Great Britain -- Foreign relations -- France -- 1648-1715
<div class="citation" vocab="http://schema.org/"><i class="fa fa-external-link-square fa-fw"></i> Data from <span resource="http://link.massey.ac.nz/resource/B2JHv_y22PQ/" typeof="CategoryCode http://bibfra.me/vocab/lite/Temporal"><span property="name http://bibfra.me/vocab/lite/label"><a href="http://link.massey.ac.nz/resource/B2JHv_y22PQ/">1648-1715</a></span> - <span property="potentialAction" typeOf="OrganizeAction"><span property="agent" typeof="LibrarySystem http://library.link/vocab/LibrarySystem" resource="http://link.massey.ac.nz/"><span property="name http://bibfra.me/vocab/lite/label"><a property="url" href="http://link.massey.ac.nz/">Massey University Library, University of New Zealand</a></span></span></span></span></div>
Data Citation of the Temporal 1648-1715
http://link.massey.ac.nz/resource/B2JHv_y22PQ/
http://library.link/resource/B2JHv_y22PQ/


On this date in 2003, Samer Ufi and Mohamed Faraj (some sources give the latter's name as Suleiman) were publicly shot by masked al-Aqsa gunmen in the West Bank town of Tulkarem (or Tulkarm) for Israeli collaboration.
A videotape of the two admitting to supplying Israel information which led to militants' assassination was played in the camp on the eve of their shooting. The dead men's families contended that they had been tortured into the confession.
Tulkarem in 2003 was a place easy to feel under siege.
Recently prosperous, the fertile district close upon the Israeli border was suffering the effects of the ongoing Palestinian rising.
Tulkarem was in the process of being riven by Israel's "apartheid wall" — splintering communities and devastating a recently prosperous economy.
Some eighteen to twenty-two thousand Palestinian laborers from the Tulkarm district used to go and work in Israel every day. Now they are prevented by the security barrier that went up during 2003 … an eight-meter-high concrete wall complete with round gray watchtowers, built to prevent Palestinian snipers from shooting at passing cars on the Trans-Israel Highway that skirts Tulkarm to the west. Additional stretches of fence hermetically seal the surrounding villages off from Israel, as well as from some of their agricultural land.
Meanwhile, as elsewhere in the West Bank and Gaza, a a fast-growing list of assassinations struck militants in the community.
We don't know in these parts whether the executed men truly were informers, but Israel is known to obtain many such targets by way of informers — often reluctant Palestinians it blackmails or bribes. Accused informers are regularly executed in the Palestinian territories.
"For myself, if I were Palestinian, I would hate them to death," an Israeli intelligence advisor told the BBC of the collaborators recruited by Tel Aviv. "He is a traitor — I need him — but he's a traitor".


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eORDER option allows eMAM users to place orders for all the assets available in the eBIN.
eMAM Users can enter the order information and comments, and this information is sent to the email address configured in the system settings. Admin can view all the eOrders made under a particular unit however regular users can only view the eOrders made by them.
Currently, this functionality is not integrated with payment gateway and the order has to be fulfilled manually.


text:
#!/usr/bin/env node

'use strict';

var commander = require('commander');
var pkg = require('../package.json');

process.title = 'soybean';

commander
	.version(pkg.version)
	.command('setup [env]')
	.description('run setup commands for all envs')
	.option("-s, --setup_mode [mode]", "Which setup mode to use")
	.action(function(env, options){
		var mode = options.setup_mode || "normal";
		env = env || 'all';
		console.log('setup for %s env(s) with %s mode', env, mode);
	})
	.on('--help', function() {
		console.log('  Examples:');
		console.log();
		console.log('    $ deploy exec sequential');
		console.log('    $ deploy exec async');
		console.log();
	});

commander
	.command('exec <cmd>')
	.alias('ex')
	.description('execute the given remote cmd')
	.option("-e, --exec_mode <mode>", "Which exec mode to use")
	.action(function(cmd, options){
		console.log('exec "%s" using %s mode', cmd, options.exec_mode);
	})
	.on('--help', function() {
		console.log('  Examples:');
		console.log();
		console.log('    $ deploy exec sequential');
		console.log('    $ deploy exec async');
		console.log();
	});

commander
	.on('--help', function() {
		console.log('  Examples:');
		console.log('    $ deploy exec sequential');
		console.log('    $ deploy exec async');
		console.log();
	});

commander.parse(process.argv);

if (!commander.args.length) commander.help();

/*
- server mode
$ soybean listen --port 10989
$ soybean user create [userid] <password>
$ soybean list [userid]
$ soybean info [userid] [instancename]
$ soybean create [userid] [instancename] <boxname>
$ soybean rm [userid] [instancename]
$ soybean start [userid] [vm]
$ soybean box list
$ soybean box add [boxname] [url]
$ boybean box remove [boxname]

- hub mode

- client global
$ soybean login
$ soybean list
$ soybean rm [instancename]

- client in project dir
$ soybean init
$ soybean start
$ soybean stop
$ soybean reload
$ soybean sync
$ soybean ssh
*/

redpajama_set_name: RedPajamaGithub


text:
Maryland State Police Investigating Motor Vehicle Collision in Great Mills
On Thursday, December 24, 2020, at approximately 12:00 p.m., police, fire and rescue personnel responded to the area of Flat Iron Road and Ryan Lane in Great Mills, for the motor vehicle collision with one subject unconscious. Troopers from the Maryland State Police Leonardtown Barrack arrived on the scene to find a single vehicle in driveway of a private residence that had struck a wooden walkway and sign belonging to Bowles Mulch, the operator was found unconscious and unresponsive inside of the vehicle. Fire and rescue personnel arrived on the scene and reported State Troopers were performing CPR on the operator. After approximately three minutes, medical personnel reported CPR was no longer in progress, with police administering at least three doses of Naloxone (NARCAN) Maryland State Police Helicopter Trooper 7 was requested and landed at a nearby landing zone, they were later cancelled and placed into service. The operator was transported to an area hospital with injuries believed to be non-life-threatening. Police are investigating the collision. Updates will be provided when they become available.
The post Maryland State Police Investigating Motor Vehicle Collision in Great Mills appeared first on Southern Maryland News Net.
Published in All News, County, Fire & Rescue, Law Enforcement, More News, St. Mary's News, z 600X120 Top Ad Bottom and z Police Ad Top
More from z Police Ad TopMore posts in z Police Ad Top »
Detectives in Charles County Seek Public's Help in Search for Fugitive / Crime Solvers Offering Cash Reward

redpajama_set_name: RedPajamaCommonCrawl


text:
layout: news
title: Accommodation pairing forum
snippet: Accommodation pairing forum is available
---

There are many [accommodation options]({{ site.github.url }}/attendees/accommodation/) available to book, but would like to link up with another attendee to share accomodation costs,
please use the [NLP roommate matching forum](https://groups.google.com/forum/#!forum/nlp-roommates).


Right Honorable Prime Minister of Cook Islands, Excellencies Former Heads of States, Honorable Ministers, Director General Dr Kande Yumkella, Eminent Scientists and Experts, Distinguished Leaders of Companies, Distinguished Guests, Ladies and Gentlemen, It is an honour for me to address this very important meeting. Until recently, energy has been one of the neglected issues in international development. We are now more aware than ever that sustainable energy systems are crucial to human livelihoods as well as ecological balance. In the developing countries, 1.6 billion people do not have access to electricity, 2.5 billion use traditional biomass for cooking while access to electricity remains very limited in rural and peri-urban areas despite the potential offered by the various sources of renewable energy. We at the ACP believe that access to renewable energy and modern energy services is a prerequisite for economic growth, social development and the attainment of the Millennium Development Goals (MDGs) and the graduation of LDCs to non LDCs status. Placing emphasis on sustainable energy is a means of consolidating the efforts made to attain the Millennium Development Goals and scale up the development of LDCs. Because of the cross-cutting or transversal nature of energy, we are attempted to agree with those who propose it as MDG Goal 9. In order to address the energy needs of ACP member states, the ACP Group and its traditional partner, the European Union agreed to put in place the 1st ACP-EU Energy Facility in 2004 with a budget of 220 Million Euros. A 2nd ACP-EU Energy Facility was launched in 2009 with a budget of 200 Million Euros. The specific objectives of the ACP-EU Energy Facility were to improve access to modern energy services for the benefit of disadvantaged rural and peri-urban population, improve governance and management in assisting countries to improve their institutional and regulatory framework to attract financial resources for sustainable energy related project. During the 1st Energy Facility, 74 projects across the ACP countries were funded. Implementation of these projects involved all categories of actors: local government, NGOs, private sector, public institutions and international organizations. The programme enabled actors in the field to explore innovative solutions that target local needs. Ladies and Gentlemen, In 2010, our principal development partner, the European Commission, released a 'Green Paper' outlining a new framework for development thinking and action. The Green Paper commits to targeting sustainable and inclusive growth as the cornerstone of the Commission's development interventions. This would imply commitment to actions that address the multi-dimensional aspects of sustainable human development, from energy to food security health, conflict and Climate Change. There is a strong emphasis on access to energy as a major focus for future development action which we have welcomed. The overall aim is to ensure that every euro spent provides the best value added, the best leverage and the best legacy of opportunities for generations to come. In our engagement with the international development community, we have continued to emphasis the increasing need to refocus attention on this crucially important question of energy as a vehicle for promoting inclusive growth and as a driver for achieving durable results in the productive sector, in particular, agriculture, food security and industrial development. ACP is aware of the immense need of the poor communities in the rural and peri-urban areas for access to sustainable energy services. But clearly, aid alone will not be sufficient to finance the investments required to address these challenges. Participation by the private sector, ACP regional development banks, as well as other financial institutions and stakeholders, would be welcome. It is against this back drop that ACP Secretariat and UNIDO decided on 24 March 2011, in Brussels to enter into a Joint Memorandum of Understanding with a view to promoting renewable sources of energy, particularly in remote rural areas. We look forward to UNIDO to work with ACP Countries in enhancing their capacity for the efficient use and management of energy, natural resources and materials to produce less pollution and waste. In concluding, let me underline the utmost importance for us to distill a clear, actionable set of priorities that will enable us evolve more robust ways to improve the access of the rural and peri-urban poor to sustainable energy services. We congratulate the UNIDO and the Government of Austria for this brilliant initiative in putting the issue of energy consequently on the international Agenda through the Vienna Forum 2011. Ladies and Gentlemen, thank you for your kind attention.


text:
\section{Introduction}\label{sec1}

In this article, we assume all rings are commutative with non-zero identity
and all modules are unital. We will start with certain definitions and
notations that will be used throughout the paper. Let $I$ be an ideal of a
ring $R$ and $K,L$ be two submodules of $R$-module $M.$ The residual $K$ by
$L$ and $I$ is defined as $(K:_{R}L)=\{r\in R\mid rL\subseteq K\}$ and
$(K:_{M}I)=\{m\in M\mid Im\subseteq K\},$ respectively. If $I$ is a principal
ideal of $R$ generated by an element $r\in R,$ then $(K:_{M}r)$ is preferred
instead of $(K:_{M}Rr).$ By $M$-$rad(N)$, we denote the $M$-radical of $N$
which is the intersection of all prime submodules containing the submodule
$N$.

Consider a multiplicatively closed subset $S$ of a ring $R$. Recently,
$S$-versions and their features of prime and primary ideals have been
thoroughly researched in various studies. Moreover, absorbing-types of such
ideals have been proposed and investigated. From the literature, we have
gathered the following definitions that we are interested in:

\begin{enumerate}
	\item A submodule $N$ of $M$ with $(N:_{R}M)\cap S=\emptyset$ is said to be
	$S$-1-absorbing prime, if there exists a fixed element $s\in S$ whenever
	$abm\in N$ for some non-unit elements $a,b\in R$ and $m\in M$, then
	$sab\in(N:_{R}M)$ or $sm\in N$, \cite{Far}.

	\item A proper ideal $I$ of $R$ is called $1$-absorbing primary if whenever
	$abc\in I$ for some non-unit $a,b,c\in R$, then either $ab\in I$ or $c\in
	\sqrt{I}$, \cite{ece1}.

	\item A proper submodule $N$ of $M$ is called $1$-absorbing primary if
	whenever $abm\in N$ for some non-unit $a,b\in R$ and $m\in M$, then either
	$ab\in(N:_{R}M)$ or $m\in M$-$rad(N)$, \cite{Ece}.

	\item A submodule $N$ of $M$ with $(N:_{R}M)\cap S=\emptyset$ is said to be
	$S$-primary, if there exists a fixed element $s\in S$ such that whenever
	$am\in N$ for some $a\in R$ and $m\in M$, then $sa\in\sqrt{(N:_{R}M)}$ or
	$sm\in N$, \cite{Ans}.

	\item A submodule $N$ of $M$ with $(N:_{R}M)\cap S=\emptyset$ is said to be
	$S$-2-absorbing primary, if there exists a fixed element $s\in S$ such that
	whenever $abm\in N$ for some $a,b\in R$ and $m\in M$, then $sam\in N$ or
	$sbm\in N$ or $sab\in\sqrt{(N:_{R}M)}$, \cite{Naji}.
\end{enumerate}

Our purpose in this paper is to extend the notion of $S$-1-absorbing primary
ideals and introduce the concept of $S$-$1$-absorbing primary submodules. This
study is composed of five sections; in the second section, the concept of
$S$-$1$--absorbing primary submodule is defined and examined in detail.
Several characterizations of $S$-$1$--absorbing primary submodules are also
presented. We exemplify the differences between $S$-$1$--absorbing primary and
$S$-$1$--absorbing prime submodules, as well as $1$--absorbing primary
submodules. We provide some theorems to examine the relations between $S
-$1$--absorbing primary submodules and the other notions such as $S$-primary
and $1$--absorbing primary submodules.

In \cite{Lu}, it has been proposed and investigated the prime avoidance
theorem for modules. This theorem says that if a submodule $N $ of $M $ is
contained in a union of finitely many prime submodules $P_{i} $'s with
$(P_{j}:M)\nsubseteq(P_{k}:M) $ whenever $j\neq k $, then $N $ is contained in
$P_{i} $ for some $i $. This theorem has also been extended to primary
submodules in various ways \cite{AtTe, DS}. In the third section, we
provide $S$-$1$--absorbing primary avoidance theorem under the assumption that
$M $ is finitely generated multiplication.

In section 4, we examine the $S$-$1$--absorbing primary property for
idealization of $M $ in $R $. The idealization of $M $ in $R $ is the ring
$R(+)M = \{(r,m) \mid r\in R, m\in M\}$ with component-wise addition and
multiplication defined as $(r_{1},m_{1})(r_{2},m_{2})=(r_{1}r_{2},r_{1
m_{2}+r_{2}m_{1})$. For more information and the ideal structure of an
idealization of $M $, the reader may see \cite{ide}.

Let $(A,B)$ be a pair of commutative rings with unity, $f:A\rightarrow B$ be a
ring homomorphism and $J$ be an ideal of $B$. In this setting, we can consider
the following subring of $A\times B:$
\[
A\bowtie^{f}J:=\left\{ (a,f(a)+j)\mid a\in A, j\in J\right\}
\]
called the amalgamation of $A$ with $B$ along $J$ with respect to $f$. This
construction is defined and examined by D'Anna et al. \cite{DF} and then
extensively investigated in different papers \cite{MCM, MES, KMSN}. In the
last section, we study $S $-1-absorbing primary ideals of the amalgamation
$A\bowtie^{f}J.$

\section{Characterization of $S$-1-absorbing primary submodules}\label{sec2}

\begin{definition}
	Let $S$ be a multiplicatively closed subset of a ring $R$ and $M$ an
	$R$-module. We call a submodule $N$ of $M$ with $(N:_{R}M)\cap S=\emptyset$ an
	$S$-$1$--absorbing primary if there exists an (fixed) $s\in S$ such that for
	all non-unit $a,b\in R$ and $m\in M$ if $abm\in N$, then either $sab\in
	(N:_{R}M)$ or $sm\in M$-$rad(N)$. This fixed element $s\in S$ is called an
	$S$-element of $N.$
\end{definition}

In particular, if one take $M=R$ as an $R$-module, then we have the definition
of $S$-$1$--absorbing primary ideal of $R$ as following:

\begin{definition}
	Let $S$ be a multiplicatively closed subset of a ring $R$. We call a proper
	ideal $I$ of $R$ with $I\cap S=\emptyset$ an $S$-$1$--absorbing primary if
	there exists an (fixed) $s\in S$ such that whenever $abc\in I$ for some
	non-unit $a,b,c\in R$, then either $sab\in I$ or $sc\in\sqrt{I}$. This fixed
	element $s\in S$ is called an $S$-element of $I.$
\end{definition}

Let $N$ be a proper submodule of $M$. It is clear that any $S$-1-absorbing
prime submodule is a $S$-1-absorbing primary submodule and that two concepts
coincide if $M$-$rad(N)=N$. However, this implication is not reversible in
general, see the following example.

\begin{example}
	\label{e1}\textbf{(}$S$\textbf{-1-absorbing primary submodule that is not
	$S$\textbf{-1-absorbing prime)} Consider the

	\mathbb{Z}

	$-module $M

	\mathbb{Z}

	\time

	\mathbb{Z}

	$ and the submodule $N=

	\mathbb{Z}

	\times

	\mathbb{Z}

	$ of $M$. For the multiplicatively closed subset $S=\{3^{n}:n>0\}$, $N$ is an
	$S$-1-absorbing primary submodule which is not $S$-1-absorbing prime.
\end{example}

\begin{proof}
	Suppose that $ab(m_{1},m_{2})\in N$ for some integers $m_{1},m_{2}$ and
	$a,b\neq\pm1$. If $m_{2}$ is even, then $sm\in

	\mathbb{Z}

	\times

	\mathbb{Z}

	=M$-$rad(N)$ for all $s\in S$. If $m_{2}$ is odd, then since $abm_{2}\in

	\mathbb{Z}

	$, we conclude that $ab\in

	\mathbb{Z}

	$, and so $sab\in2

	\mathbb{Z}

	=(N:_{R}M)$ for all $s\in S.$ Thus $N$ is an $S$-1-absorbing primary
	submodule. As $2\cdot3\cdot(1,4)\in N$ but neither $6s\in(N:_{R}M)=2

	\mathbb{Z}

	$ nor $s(1,4)\in N$ for any $s\in S$, $N$ is not $S$-1-absorbing prime.
\end{proof}

Let $S$ be a multiplicatively closed subset of a ring $R.$ In the following
proposition, by $S^{\ast},$ we denote the saturation set of $S$ defined by
$\{r\in R:\frac{r}{1}\in U(S^{-1}R)\}.$ It is clear that $S\subseteq S^{\ast}$
and $S^{\ast}$ is also a multiplicatively closed subset of $R$. In case of
$S=S^{\ast}$, then $S$ is said to be a saturated set, \cite{Gilmer}. For an
$R$-module $M,$ by $U_{M}(R)$, we denote the set $\{r\in R:rM=M\}$ which
contains $U(R)$ is a saturated multiplicatively closed subset of $R$.

\begin{proposition}
	\label{p1}Let $S$ be a multiplicatively closed subset of a ring $R$, and $N$
	be a submodule of an $R$-module $M$.
\end{proposition}

\begin{enumerate}
	\item Let $T\subseteq S$ be two multiplicatively closed subsets of $R.$ If $N$
	is an $T$-1-absorbing primary submodule such that $(N:_{R}M)\cap S=\emptyset$,
	then $N$ is also an $S$-1-absorbing primary submodule. Conversely, if for each
	$s\in S$, there is an element $s^{\prime}\in T$ such that $ss^{\prime}\in T$
	and $N$ is a $S$-1-absorbing primary submodule of $M$, then $N$ is an
	$T$-1-absorbing primary submodule of $M$.

	\item Every $1$-absorbing primary submodule $N$ with $(N:_{R}M)\cap
	S=\emptyset$, is $S$-1-absorbing primary. However, it is clear that these two
	classes of $1$-absorbing primary and $S$-1-absorbing primary submodules
	coincide if $S\subseteq U(R).$

	\item $N$ is an $S$-1-absorbing primary\ submodule of $M$ if and only if $N$
	is $S^{\star}$-1-absorbing primary.

	\item Suppose that $M=Rm$ is a cyclic $R$-module and $S\subseteq U_{M}(R).$
	Then $N$ is a 1-absorbing primary submodule if and only if $N$ is an
	$S$-1-absorbing primary submodule.
\end{enumerate}

\begin{proof}
	(1) It is clear.

	(2) The claim follows by taking $T=\{1\}$ in (1).

	(3) Suppose that $N$ is $S$-1-absorbing primary submodule. It is clear that
	$(N:_{R}M)\cap S^{\ast}=\emptyset$. Since $S\subseteq S^{\star}$, then by (1),
	$N$ is $S^{\star}$-1-absorbing primary. For the converse, let $abm\in N$ for
	some non-unit elements $a,b\in R$ and $m\in M$. Since $N$ is an $S^{\star
	$-1-absorbing primary submodule, there exists an $s\in S^{\star}$ such that
	$sab\in(N:_{R}M)$ or $sm\in M$-$rad(N)$. Since $\frac{s}{1}$ is a unit of
	$S^{-1}R$, $sur\in S$ for some $u\in S$ and $r\in R$. Say $t:=sur\in S.$
	Hence, we conclude $tm\in M$-$rad(N)$ or $tab\in(N:_{R}M)$, as needed.

	(4) Suppose that $N$ is an $S$-1-absorbing primary submodule. Let
	$abm^{\prime}\in N$ for some non-unit $a,b\in R$ and $m^{\prime}=rm\in M$.
	Then there exists an $s\in S$ such that $sab\in(N:_{R}M)$ or $srm\in
	M$-$rad(N).$ If $srm\in M$-$rad(N)$, then since $S\subseteq U_{M}(R)$ we have
	$srRm=rM\subseteq M$-$rad(N)$ which implies $m^{\prime}=rm\in M$-$rad(N)$. If
	$sab\in(N:_{R}M)$, then since $M$ is a multiplication module and $S\subseteq
	U_{M}(R),$ we conclude $abM=sabM\subseteq(N:_{R}M)M\subseteq M$-$rad(N).$ Thus,
	$N$ is a 1-absorbing primary submodule. The converse part follows from (2).
\end{proof}

\begin{example}
	\label{e2}\textbf{(}$S$\textbf{-1-absorbing primary submodule that is not
		1-absorbing primary)} Consider the multiplicatively closed subset
	$S=\{2^{k}3^{t}:$ $k,t\geq0\}\ $of

	\mathbb{Z}

	.$ Then the

	\mathbb{Z}

	$-submodule $N=(0)$ of $M

	\mathbb{Q}

	\time

	\mathbb{Z}

	_{2}\time

	\mathbb{Z}

	_{3}$ is an $S$-1-absorbing primary submodule which is not 1-absorbing primary.
\end{example}

\begin{proof}
	Suppose that $ab\left(  \frac{c}{d},\overline{x},\overline{y}\right)  \in N$
	for some $\pm1\neq a,b\i

	\mathbb{Z}

	$ and $\left(  \frac{c}{d},\overline{x},\overline{y}\right)  \in M$. Consider
	the element $6\in S.$ Here, $abc=0$ which implies $a=0$ or $b=0$ or $c=0$. If
	$a=0$ or $b=0$, then clearly $sab\in(N:_

		\mathbb{Z}

	}M)$. If $c=0,$ then $6\left(  \frac{c}{d},\overline{x},\overline{y}\right)
	=(0,\overline{0},\overline{0})\in M$-$rad(N)=(0)$, and thus $N$ is an
	$S$-1-absorbing primary submodule. However, it is not $1$-absorbing primary as
	$2\cdot3\cdot(0,\overline{1},\overline{1})\in N$ but neither $2\cdot3\in(N:_

		\mathbb{Z}

	}M)=0$ nor $(0,\overline{1},\overline{1})\in M$-$rad(N)=(0)$.
\end{proof}

\begin{theorem}
	\label{char}Let $S$ be a multiplicatively closed subset of a ring $R.$ For a
	submodule $N$ of an $R$-module $M$ satisfying $(N:_{R}M)\cap S=\emptyset$, the
	following are equivalent.
\end{theorem}

\begin{enumerate}
	\item $N$ is an $S$-$1$-absorbing primary submodule of $M$.

	\item There exists an $s\in S$ such that whenever $Ibm$ $\subseteq N$ for some
	proper ideal $I$ of $R,$ $b\in R\backslash U(R)$ and $m\in M$ implies either
	$sIb\subseteq\left(  N:_{R}M\right)  $ or $sm\in M$-$rad(N)$.

	\item There exists an $s\in S$ such that whenever $IJm$ $\subseteq N$ for some
	proper ideals $I$, $J$ of $R$ and $m\in M$ implies either $sIJ\subseteq\left(
	N:_{R}M\right)  $ or $sm\in M$-$rad(N)$.

	\item There exists an $s\in S$ such that whenever $IJK$ $\subseteq N$ for some
	proper ideals $I$, $J$ of $R$ and a submodule $K$ of $M$ implies either
	$sIJ\subseteq\left(  N:_{R}M\right)  $ or $sK\subseteq M$-$rad(N)$.
\end{enumerate}

\begin{proof}
	(1)$\Rightarrow$(2). Let $s$ be an $S$-element of $N$. Suppose that $Ibm$
	$\subseteq N$ and $sIb\nsubseteq\left(  N:_{R}M\right)  $. Then $sxb\notin
	\left(  N:_{R}M\right)  $ for some non-unit $x\in I$. Since $xbm\in N$ and
	$sxb\notin\left(  N:_{R}M\right)  ,$ we have $sm$ $\in M$-$rad(N),$ as needed.

	(2)$\Rightarrow$(3). Let $s$ be the fixed element in (2). Suppose that $IJm$
	$\subseteq N$ but $sIJ\nsubseteq\left(  N:_{R}M\right)  $ for some proper
	ideals $I$, $J$ of $R$ and $m\in M$. Then $sIb\notin\left(  N:_{R}M\right)  $
	for some non-unit $b\in J$. From our assumption (2), we conclude $sm$ $\in
	M$-$rad(N)$.

	(3)$\Rightarrow$(4). Let $s$ be the fixed element in (3). Suppose $IJK$
	$\subseteq N$ but $sK$ $\nsubseteq N$ for some proper ideals $I$, $J$ of $R$
	and a submodule $K$ of $M.$ Hence $sm\notin M$-$rad(N)$ for some $m\in K$ which
	implies $sIJ\subseteq\left(  N:_{R}M\right)  $ by (3).

	(4)$\Rightarrow$(1). Let $a,b\in R$ be non-unit elements, $m\in M$ and $abm\in
	N$. Put $I$ $=aR$, $J=bR$ and $K=Rm$ in (4), then we are done.
\end{proof}

In the following, we characterize $S$-$1$-absorbing primary ideals of a ring
$R$ by taking $M=R$ as an $R$-module in Theorem \ref{char}.

\begin{theorem}
	\label{char1}Let $S$ be a multiplicatively closed subset of a ring $R$ and $I$
	an ideal of $R$ with $I\cap S=\emptyset$. Then $I$ is an $S$-$1$-absorbing
	primary ideal of $R$ if and only if there exists an $s\in S$ such that
	whenever $I_{1}I_{2}I_{3}\subseteq I$ for some proper ideals $I_{1},I_{2}$ and
	$I_{3}$ of $R$ implies either $sI_{1}I_{2}\subseteq I$ or $sI_{3
	\subseteq\sqrt{I.}$
\end{theorem}

Recall from \cite[Lemma 2.4]{Hoj} that if $M$ is a finitely generated
multiplication $R$-module, then $\sqrt{(N:_{R}M)}=(M$-$rad(N):_{R}M)$ for any
submodule $N$ of $M$

Next, we give a characterization for $S$-$1$-absorbing primary submodules in
terms of submodules of a finitely generated multiplication modules.

\begin{theorem}
	Let $S$ be a multiplicatively closed subset of a ring $R$ and $M$\ be a
	finitely generated multiplication $R$-module. A proper submodule $N$ of $M$
	with $(N:_{R}M)\cap S=\emptyset$ is an $S$-$1$-absorbing primary submodule of
	$M$ if and only if there exists an $s\in S$ such that whenever $N_{1
	N_{2}N_{3}\subseteq N$ for some proper submodules $N_{1}$, $N_{2},$ $N_{3}$ of
	$M$ implies either $sN_{1}N_{2}\subseteq N$ or $sN_{3}\subseteq M$-$rad(N)$.
\end{theorem}

\begin{proof}
	Suppose that $N_{1}N_{2}N_{3}$ $\subseteq N$ for some proper submodules
	$N_{1}=I_{1}M,$ $N_{2}=I_{2}M$ and $N_{3}=I_{3}M$ where $I_{1}$, $I_{2},$
	$I_{3}$ are proper ideals of $R$. Hence $I_{1}I_{2}I_{3}\subseteq(N:_{R}M)$
	and $(N:_{R}M)$ is a $S$-$1$-absorbing primary ideal of $R$ by Lemma \ref{d}.
	This yields that there exists $s\in S$ such that $sI_{1}I_{2}\subseteq\left(
	N:_{R}M\right)  $ or $sI_{3}\subseteq\sqrt{(N:_{R}M)}$ by Theorem \ref{char1}.
	Thus we conclude either $sN_{1}N_{2}=sIJM\subseteq\left(  N:_{R}M\right)  M=N$
	or $sN_{3}\subseteq\sqrt{(N:_{R}M)}M=M$-$rad(N)$, as desired. For the converse,
	suppose that $IJm$ $\subseteq N$ for some proper ideals $I$, $J$ of $R$ and
	$m\in M$. Put $N_{1}=IM$, $N_{2}=JM$ and $N_{3}=Rm$. This follows that
	$sIJ\subseteq(N:_{R}M)$ or $sm\in sN_{3}\subseteq M$-$rad(N)$, thus we are done
	by Theorem \ref{char}.
\end{proof}

Now, we are ready for a characterization of $S$-1-absorbing primary submodules
of finitely generated faithful multiplication modules in terms of presentation
ideals of them.

\begin{theorem}
	\label{char2}Let $M$ be a finitely generated faithful multiplication
	$R$-module and $S$ be a multiplicatively closed subset of $R.$ For a submodule
	$N$ of $M$ provided $(N:_{R}M)\cap S=\emptyset,$ the following statements are equivalent.
\end{theorem}

\begin{enumerate}
	\item $N$ is a $S$-1-absorbing primary submodule of $M$.

	\item $N=IM$ for some $S$-1-absorbing primary ideal $I$ of $R$.

	\item $(N:_{R}M)$ is an $S$-1-absorbing primary ideal of $R$.
\end{enumerate}

\begin{proof}
	(1)$\Rightarrow$(2)$.$ Let $N$ be a $S$-1-absorbing primary submodule of $M$.
	Then $(N:_{R}M)$ is an $S$-1-absorbing primary ideal of $R$ by Proposition
	\ref{d}. Since $M$ is multiplication, put $I=(N:_{R}M)$, so the claim is clear.

	(2)$\Rightarrow$(3)$.$ Since $M\ $is cancellation, that is,
	$IM=JM\Leftrightarrow I=J$ for every ideal $I,J\ $of $R$ which means that
	$(N:_{R}M)=(IM:_{R}M)=I$ is an $S$-1-absorbing primary ideal of $R$.

	(3)$\Rightarrow$(1)$.$ Suppose that $0\neq IJK\subseteq N\ $for some proper
	ideals $I,J\ $of $R\ $and a submodule $K\ $of $M.\ $Since $M$ is a
	multiplication module, $K=LM\ $for some ideal $L\ $of $R.\ $Then we have
	$IJLM\subseteq(N:_{R}M).\ $We may assume that $L\ $is a proper ideal of
	$R.\ $Since $(N:_{R}M)\ $is an $S$-1-absorbing primary ideal of $R,$ so it is
	an $S$-1-absorbing primary submodule of an $R$-module $R$ and there exists a
	fixed $s\in S$ such that $sIJ\subseteq(N:_{R}M)$ or $sL\subseteq\sqrt
	{(N:_{R}M)}=(M$-$rad(N):_{R}M)$ by Theorem \ref{char1}. This yields that
	$sIJ\subseteq(N:_{R}M)$ or $sK=LM\subseteq M$-$rad(N)$ and $N\ $is
	$S$-1-absorbing primary submodule of $M$ by Theorem \ref{char}.$\ $
\end{proof}

Every $S$-primary submodule of a finitely generated faithful multiplication module is $S$-1-absorbing primary, see the following.

\begin{proposition}
	Let $S$ be a multiplicatively closed subset of a ring $R$ and $M$ be a
	finitely generated faithful multiplication $R$-module. If $N$ is an
	$S$-primary submodule of $M,$ then it is $S$-1-absorbing primary.
\end{proposition}

\begin{proof}
	Let $N$ be an $S$-primary submodule of $M.$ Then $(N:_{R}M)$ is a $S$-primary
	ideal of $R$ by \cite[Proposition 2.11 (1)]{Ans} and hence $(N:_{R}M)$ is a
	$S$-1-absorbing primary ideal of $R.$ Thus $N$ is an $S$-1-absorbing primary
	submodule of $M$ by Theorem \ref{char2}.
\end{proof}

\begin{lemma}
	\label{lrad}Let $S$ be a multiplicatively closed subset of $R.$ If $I$ is an
	$S$-1-absorbing primary ideal, then $\sqrt{I}$ is an $S$-prime ideal of $R$.
\end{lemma}

\begin{proof}
	Let $s$ be an $S$-element of $I$, $a,b\in R$ and $ab\in\sqrt{I}$. If $a$ or
	$b$ is unit, then the claim is clear. So, assume that $a,b$ are non-unit
	elements. Then $a^{k}b^{k}=a^{t}a^{t}b^{k}\in I$ for some $k=2t>1$ which
	implies that $sa^{k}\in I$ or $sb^{k}\in\sqrt{I}$. Thus $sa\in\sqrt{I}$ or
	$sb\in\sqrt{I}$ and $\sqrt{I}$ is an $S$-prime ideal of $R$.
\end{proof}

\begin{proposition}
	Let $S$ be a multiplicatively closed subset of $R.$ If $N$ is an
	$S$-1-absorbing primary submodule of $M$, then $M$-$rad(N)$ is an $S$-prime
	submodule of $M$.
\end{proposition}

\begin{proof}
	Follows from Theorem \ref{char2}, Lemma \ref{lrad} and \cite[Proposition 2.9
	(2)]{Sev}.
\end{proof}

Let $S$ be a multiplicatively closed subset of a ring $R$. Then $S$ is said to
satisfy the maximal multiple condition if there exists $s\in S$ such that $t$
divides $s$ for each $t\in S$, \cite{And}. For instance, if $S$ is finite,
then $S$ satisfies the maximal multiple condition. Also, $U(R)$ is a
multiplicatively closed set satisfying the maximal multiple condition.

\begin{proposition}
	Let $S$ be a multiplicatively closed subset of a ring $R$, and $N$ be a
	submodule of an $R$-module $M$.
\end{proposition}

\begin{enumerate}
	\item If $N$ is an $S$-1-absorbing primary submodule of $M$, then $S^{-1}N$ is
	a 1-absorbing primary $S^{-1}R$-submodule of $S^{-1}M$.

	\item If $S$ satisfies the maximal multiple condition, and $S^{-1}N$ is a
	1-absorbing primary $S^{-1}R$-submodule of $S^{-1}M$, then $N$ is an
	$S$-1-absorbing primary submodule of $M.$
\end{enumerate}

\begin{proof}
	(1) Let $\frac{a}{s_{1}},\frac{b}{s_{2}}$ be non-unit elements of $S^{-1}R$
	and $\frac{m}{s_{3}}\in S^{-1}M$ with $\frac{a}{s_{1}}\frac{b}{s_{2}}\frac
	{m}{s_{3}}\in S^{-1}N$. Then $uabm\in N$ for some $u\in S$. Hence, there
	exists an $s\in S$ such that $suab\in(N:_{R}M)$ or $sm\in M$-$rad(N)$. This
	yields either $\frac{a}{s_{1}}\frac{b}{s_{2}}=\frac{suab}{sus_{1}s_{2}}\in
	S^{-1}(N:_{R}M)\subseteq(S^{-1}N:_{S^{-1}R}S^{-1}M)$ or $\frac{m}{s_{3}
	=\frac{sm}{ss_{3}}\in S^{-1}(M$-$rad(N))=S^{-1}M$-$rad(S^{-1}N)$ and $S^{-1}N$ is
	a 1-absorbing primary submodule of $S^{-1}M$.

	(2) Let $s_{\max}\in S$ is the element which has the maximal multiple
	condition. Suppose that $abm\in N$ for some non-unit elements $a,b\in R$ and
	$m\in M$. Then $\frac{a}{1}\frac{b}{1}\frac{m}{1}\in S^{-1}N$ which implies
	$\frac{a}{1}\frac{b}{1}\in(S^{-1}N:_{S^{-1}R}S^{-1}M)\subseteq$ $S^{-1
	(N:_{R}M)$ or $\frac{m}{1}\in S^{-1}M$-$rad(S^{-1}N)=S^{-1}(M$-$rad(N))$. This
	follows that $uab\in(N:_{R}M)$ or $vm\in M$-$rad(N)$. Since $u$ and $v$ divide
	$s_{\max}$, we get $s_{\max}ab\in(N:_{R}M)$ or $s_{\max}m\in M$-$rad(N)$ as needed.
\end{proof}

\begin{proposition}
	Let $S$ be a multiplicatively closed subset of $R,$ and $M$ be a
	multiplication $R$-module. If $\{N_{i}\}_{i=1}^{k}$ is a family of
	$S$-1-absorbing primary submodules of $M$ with the same $M$-radical, then

	{\displaystyle\bigcap\limits_{i=1}^{k}}

	N_{i}$ is also $S$-1-absorbing primary submodule of $M.$
\end{proposition}

\begin{proof}
	Let $s_{i}\in S$ be an $S$-element of $N_{i}$ for each $i=1,...,k.$ First,
	note that $\left(

	{\displaystyle\bigcap\limits_{i=1}^{k}}

	N_{i}\right)  \cap S=\emptyset$. Put $s=s_{1}s_{2}...s_{k}$. Suppose that
	$abm\i

	{\displaystyle\bigcap\limits_{i=1}^{k}}

	N_{i}$ but $sab\notin\left(

	{\displaystyle\bigcap\limits_{i=1}^{k}}

	N_{i}:_{R}M\right)  $ for some non-unit elements $a,b\in R$ and $m\in M$. Then
	$sab\notin(N_{j}:_{R}M)$ for some $j\in\{1,\dots,k\}.$ Since $N_{j}$ is
	$S$-1-absorbing primary and $abm\in N_{j}$, we have $s_{j}m\in M$-$rad(N_{j})$
	and so $sm\in M$-$rad(N_{j})=M$-$rad\left(

	{\displaystyle\bigcap\limits_{i=1}^{k}}

	N_{i}\right)  $ by \cite[Proposition 2.14 (3)]{Hoj}, we are done.
\end{proof}

\begin{lemma}
	\label{lemf}Let $f:M_{1}\rightarrow M_{2}$ be an $R$-module epimorphism and
	$S$ a multiplicatively closed subset of $R$.
\end{lemma}

\begin{enumerate}
	\item If $N$ is a submodule of $M_{1}$, then $(N:_{R}M_{1})\subseteq
	(f(N):_{R}M_{2}),$ \cite{Ece}.

	\item If $K$ is a submodule of $M_{2}$, then$(K:_{R}M_{2})\subseteq
	(f^{-1}(K):_{R}M_{1})$, \cite{Ece}.

	\item If $N$ is a submodule of $M_{1}$ and $\ker\left(  f\right)  \subseteq
	N$, then $f\left(  M_{1}\text{-}rad(N)\right)  =M_{2}$-$rad(f\left(  N\right)
	)$, \cite{Lu2}.

	\item If $K$ is a submodule of $M_{2}$, then $f^{-1}\left(  M_{2
	\text{-}rad(K)\right)  =M_{1}$-$rad(f^{-1}\left(  K\right)  )$, \cite{Lu2}.
\end{enumerate}

\begin{proposition}
	\label{f}Let $f:M_{1}\rightarrow M_{2}$ be an $R$-module epimorphism and $S$
	be a multiplicatively closed subset of $R$.
\end{proposition}

\label{p/}

\begin{enumerate}
	\item Suppose that $\left(  f\left(  N\right)  :_{R}M_{2}\right)  \cap
	S=\emptyset$. If $N$ is an $S$-1-absorbing primary submodule of $M_{1}$
	containing $Ker(f)$, then $f(N)$ is an $S$-1-absorbing primary submodule of
	$R_{2}.$

	\item If $K$ is an $S$-1-absorbing primary submodule of $M_{2}$, then
	$f^{-1}(K)$ is an $S$-1-absorbing primary submodule of $M_{1}.$
\end{enumerate}

\begin{proof}
	(1) Let $a,b$ be non-unit elements of $R$ and $m\in M$ with $abm\in f(N)$. Say
	$m=f(m_{1})$ for some $m_{1}\in M_{1}$. Since $abf(m_{1})\in f(N)$ and
	$Ker(f)\subseteq I$, we have $abm_{1}\in N$ which implies that there exists an
	$s\in S$ such that $sab\in(N:_{R}M_{1})$ or $sm\in M_{1}$-$rad(N)$. Thus,
	$sab\in(N:_{R}M_{1})\subseteq f((N):_{R}M_{2})$ or $sf(m)\in f(N)\subseteq
	M_{2}$-$rad(f(N))$ by Lemma \ref{lemf}.

	(2) First, we show that $\left(  f^{-1}\left(  K\right)  :_{R}M_{1}\right)
	\cap S=\emptyset.$ Assume that $r\in\left(  f^{-1}\left(  K\right)  :_{R
	M_{1}\right)  \cap S$. Then $rM_{1}\subseteq f^{-1}\left(  K\right)  $ implies
	$rM_{2}=rf(M_{1})\subseteq f(f^{-1}\left(  K\right)  )\subseteq K$, and hence
	$r\in(K:_{R}M_{2})$, a contradiction. Let $a$ and $b$ be non-unit elements of
	$R$ and $m\in M_{1}$ with $abm\in f^{-1}(K).$ Then $abf(m)\in K$ which implies
	that there exists $s\in S$ such that $sab\in(K:_{R}M_{2})$ or $sf(m)\in
	M_{2}$-$rad(K)$. Thus, we conclude $sab\in(K:_{R}M_{2})\subseteq(f^{-1
	(K):_{R}M_{1})$ or $sm\in f^{-1}\left(  M_{2}\text{-}rad(K)\right)
	=M_{1}$-$rad(f^{-1}(K))$ by Lemma \ref{lemf}. Therefore, $f^{-1}(K)$ is an
	$S$-1-absorbing primary submodule of $M_{1}.$
\end{proof}

As a consequence of Theorem \ref{f}, we have the following result.

\begin{corollary}
	\label{c/}Let $S$ be a multiplicatively closed subset of $R$ and $N,$ $K$ be
	submodules of $R$-module $M$.
\end{corollary}

\begin{enumerate}
	\item Let $K\subseteq N$. Then $N$ is an $S$-1-absorbing primary submodule of
	$M$ if and only if $N/K$ is an $S$-1-absorbing primary submodule of $M/K$.

	\item If $N$ is an $S$-1-absorbing primary submodule of $M$ with
	$(N:_{R}K)\cap S=\emptyset$, then $N\cap K$ is an $S$-1-absorbing primary
	submodule of $K$.
\end{enumerate}

\begin{proof}
	(1) It is clear that $\left(  N/K:_{R}M/K\right)  \cap S=\emptyset$. Consider
	the canonical epimorphism $\pi:M\rightarrow M/K$ in Proposition \ref{f} (1).
	Then $N/K$ is an $S$-1-absorbing primary submodule of $M/K$. For the converse
	part, suppose that $abm\in N$ for some non-unit elements $a$ and $b$ of $R$,
	$m\in M.$ Hence $ab(m+K)\in N/K$ which yields that there exists a fixed $s\in
	S$ such that $sab\in(N/K:_{R}M/K)$ or $s(m+K)\in M/K$-$rad(N/K)=M$-$rad(N)/K.$
	Therefore $sab\in(N:_{R}M)$ or $sm\in M$-$rad(N),$ so we are done.

	(2) Consider that the injection $i:K\rightarrow M$ defined by $i(k)=k$ for all
	$k\in K$. It is easy to see that $(i^{-1}(N):K)\cap S=\emptyset$. Then
	$i^{-1}(N)=N\cap K$ is is an $S$-1-absorbing primary submodule of $K$ by
	Proposition \ref{f} (2).
\end{proof}

Next, we show that if a ring $R$ admits an $S$-1-absorbing primary ideal that
is not an $S$-primary ideal, then $R$ is a quasilocal ring. First, recall the
following lemma.

\begin{lemma}
	\cite{ece1}\label{Lq} Let $R$ be a ring. Suppose that for every non-unit
	element $w$ of $R$ and for every unit element $u$ of $R$, we have $w+u$ is a
	unit element of $R$. Then $R$ is a quasilocal ring.
\end{lemma}

\begin{theorem}
	\label{Tq} Let $R$ be a ring. If there is an $S$-1-absorbing primary ideal
	which is not an $S$-primary, then $R$ is a quasilocal ring.
\end{theorem}

\begin{proof}
	Let $s^{\prime}\in S$ be an $S$-element of $J$ which is not $S$-primary ideal
	of $R$. Then there exist non-unit elements $a,b\in R$ with $ab\in J$
	satisfying $sa\notin J$ and $sy\notin\sqrt{J}$ for all $s\in S$. Choose $r$
	$\in R\backslash U(R)$. Hence $rab\in J$ which yields $s^{\prime}ra\in J$. We
	show that $r+u\in U(R)$ for any $u\in U(R)$. Let $u\in U(R)$. Assume that
	$r+u$ $\in R\backslash U(R)$. Then $(r+u)ab\in J$ and $s^{\prime}y\notin
	\sqrt{I}$ implies that $s^{\prime}(r+u)a\in I$. Now, since $s^{\prime}ra\in
	I,$ we get $s^{\prime}ua\in I$, and so $s^{\prime}a\in J$, a contradiction.
	Therefore, $r+u$ $\in U(R)$ and $R$ is a quasilocal ring by Lemma \ref{Lq}.
\end{proof}

As a direct result of Theorem \ref{Tq}, we have the following corollary.

\begin{corollary}
	\label{cq} Let $I$ be an ideal of a non-quasi local ring $R$ with $I\cap
	S=\emptyset$. Then $I$ is an $S$-1-absorbing primary ideal of $R$ if and only
	if $I$ is an $S$-primary ideal of $R$.
\end{corollary}

\begin{lemma}
	\label{d}Let $S$ be a multiplicatively closed subset of $R$. If $N$ is an
	$S$-1-absorbing primary submodule of a finitely generated multiplication
	$R$-module $M$. Then the following statements hold.
\end{lemma}

\begin{enumerate}
	\item $(N:_{R}M)$ is an $S$-1-absorbing primary ideal of $R$.

	\item If $(N:_{R}m)\cap S=\emptyset$ for some $m\in M\backslash\mathfrak{P}$
	where $\mathfrak{P}$ is any prime submodule of $M$ containing $N,$ then
	$(N:_{R}m)$ is an $S$-1-absorbing primary ideal of $R$.

	\item If $((N:_{M}r):_{R}M)\cap S=\emptyset$ for some $r\in R\backslash
	(N:_{R}M)$, then $(N:_{M}r)$ is an $S$-1-absorbing primary submodule of $M$.
\end{enumerate}

\begin{proof}
	(1) Let $a,b,c$ $\in R\backslash U(R)$ and $abc\in(N:_{R}M)$. Then
	$ab(cM)\subseteq N$ implies that there exists an $s\in S$ such that
	$sab\in(N:_{R}M)$ or $scM\subseteq M$-$rad(N)$ by Theorem \ref{char}. Thus we
	have $sab\in(N:_{R}M)$ or $sc\in (M$-$rad(N):_{R}M)  =\sqrt
	{(N:_{R}M)}$ and $(N:_{R}M)$ is an $S$-1-absorbing primary ideal of $R$.

	{(2) Let $a,b,c$ $\in R\setminus U(R)$ and $abc\in(N:_{R}m)$. Then we have
		$abcm\in N.$ As $N$ is an $S$-1-absorbing primary submodule, there exists an
		$s\in S$ such that $sab\in(N:_{R}M)$ or $scm\in M$-$rad(N).$ This indicates that
		$sab\in(N:_{R}m)$ or $sc\in(M$-$rad(N):_{R}m).$ For the rest of the proof, we
		only need to show that $(M$-$rad(N):_{R}m)=\sqrt{(N:_{R}m)}.$ The inclusion
		$\sqrt{(N:_{R}m)}\subseteq(M$-$rad(N):_{R}m)$ is satisfied by (\cite{MRS}, Lemma
		2.1). For the reverse inclusion, let $a\in(M$-$rad(N):_{R}m)$ which implies
		$am\in M$-$rad(N)=\bigcap_{N\subseteq P}P$ where $P$ is any prime submodule of
		$M$ that contains $N.$ Since $M$ is multiplication, there exists an ideal $I$
		of $R$ such that $N=IM$ and a prime ideal $Q$ of $R$ such that $P=QM.$ It
		follows that $am\in\bigcap_{N\subseteq P}P=\bigcap_{IM\subseteq QM}QM.$ Then,
		we obtain $am\in QM$ for all prime ideals $Q$ containing $I$ by using
		(\cite{El}, Theorem 3.2). This means that $a\in Q$ or $m\in QM$ for all primes
		$Q$ containing $I$ by (\cite{El}, Lemma 2.10) and hence $a\in\bigcap
		_{I\subseteq Q}Q=\sqrt{I}$ which implies $a^{n}\in I$ for some $n\in
		\mathbb{N}.$ We conclude that $a^{n}m\in IM=N$, namely $a\in\sqrt{(N:_{R}m)},$
		which completes the proof. }

	(3) Suppose that $abm\in(N:_{M}r)$ for some non-unit $a,b\in R$ and $m\in M$.
	Then $abrm\in N$ which implies that there exists an $s\in S$ such that
	$sab\in\left(  N:_{R}M\right)  \subseteq((N:_{M}r):_{R}M)$ or $srm\in
	M$-$rad(N)\subseteq M$-$rad((N:_{M}r))$. Thus $(N:_{M}r)$ is an $S$-1-absorbing
	primary submodule of $M$.
\end{proof}

\begin{theorem}
	Let $M$ be a finitely generated multiplication $R$-module where $R$ is a
	non-quasi local ring and $N$ be a submodule of $M$ with $(N:_{R}M)\cap
	S=\emptyset$. Then $N$ is an $S$-1-absorbing primary submodule of $M$ if and
	only if $N$ is an $S$-primary submodule of $M.$
\end{theorem}

\begin{proof}
	Suppose that $N$ is a $S$-1-absorbing primary submodule of $M$. Then
	$(N:_{R}M)$ is an $S$-1-absorbing primary ideal of $R$ by Lemma \ref{d} and
	$(N:_{R}M)$ is $S$-primary ideal of $R$ by Corollary \ref{cq}. Hence, $N$ is
	an $S$-primary submodule of $M$ by \cite[Theorem 2.17]{Ans}. Conversely, if
	$N$ is an $S$-primary submodule of $M,$ then $(N:_{R}M)$ is an $S$-primary
	ideal of $R$ by \cite[Theorem 2.17]{Ans}$.$ Since every $S$-primary ideal with
	$(N:_{R}M)\cap S=\emptyset$ is an $S$-1-absorbing primary ideal, $(N:_{R}M)$
	is an $S$-1-absorbing primary ideal of $R.$ Thus, $N$ is an $S$-1-absorbing
	primary submodule of $M$ by Theorem \ref{char}.
\end{proof}

In view of Corollary \ref{cq}, we have the following result.

\begin{theorem}
	Let $S_{i}$ be multiplicatively closed subset of ring $R_{i}$ for each
	$i=1,\dots,n$ and $S=S_{1}\times\cdots\times S_{n}$. For an ideal
	$I=I_{1}\times\cdots\times I_{n}$ of $R=R_{1}\times\cdots\times R_{n},$ the
	following are equivalent.

	\begin{enumerate}
		\item $I$ is an $S$-1-absorbing primary ideal of $R$.

		\item $I$ is an $S$-primary ideal of $R$.

		\item $I_{j}$ is an $S_{j}$-primary ideal of $R_{j}$ for some $j\in
		\{1,\dots,n\}$ and $I_{k}\cap S_{k}\neq\emptyset$ for all $k\in\{1,\dots
		,n\}\backslash\{j\}$.
	\end{enumerate}

	\begin{proof}
		(1)$\Leftrightarrow$(2) Since $R$ is not quasi-local, $S$-1-absorbing primary
		ideals and $S$-primary ideals coincide by Corollary \ref{cq}.
		(2)$\Leftrightarrow$(3) Follows from \cite[Theorem 2.21]{Ans} considering
		$M_{i}=R_{i}$ as an $R_{i}$-module for all $i\in\{1,\dots,n\}.$
	\end{proof}
\end{theorem}

\begin{proposition}
	Let $S$ be a multiplicatively closed subset of $R,$ $M_{1}$, $M_{2}$ be
	$R$-modules and $M=M_{1}\times M$ and $N_{1}$ be a proper submodule of $M_{1
	$. If $N=N_{1}\times M_{2}$ is an $S\times0$-1-absorbing primary submodule of
	$R$-module $M$, then $N_{1}$ is an $S$-1-absorbing primary submodule of
	$M_{1}.$
\end{proposition}

\begin{proof}
	Suppose that $N=N_{1}\times M_{2}$ is an $S\times0$-1-absorbing primary
	submodule of $M$. Then $N_{1}\cong N/(\{0\}\times M_{2})$ is an $S
	-1-absorbing primary submodule of $M/(\{0\}\times M_{2})\cong M_{1}$ by
	Corollary \ref{c/}.
\end{proof}

\begin{theorem}
	Let $S$ be a multiplicatively close subset of $R,$ $M_{1},$ $M_{2}$ be
	$R$-modules and $M=M_{1}\times M_{2}$ and $N_{1},N_{2}$ be proper submodules
	respectively of $M_{1}$ and $M_{2}.$ Let $M=M_{1}\times M_{2}$,
	$N=N_{1}\times M_{2}$ and $ N^{\prime}=M_{1}\times N_{2}.$ We suppose that
	$(N_{i}:M_{i})\cap S=\emptyset,$ where $i=1,2.$

	\begin{enumerate}
		\item $N_{1}$ is an $S$-1-absorbing primary submodule of $M_{1}$ if and only if
		$N$ is an $S\times\{1\}$-1-absorbing primary submodule of $M.$

		\item $N_{2}$ is an $S$-1-absorbing primary submodule of $M_{2}$ if and only if
		$N^{\prime}$ is a $\{1\}\times S$-1-absorbing primary submodule of $M.$
	\end{enumerate}
\end{theorem}

\begin{proof}

	(1) Suppose that $N$ is an $S$-1-absorbing primary submodule of $M,$ let
	$(s,1)$ be $S\times\{1\}$-element of $N.$ Let $a,b$ be non-nunit elements of
	$R$ and $m_{1}\in M_{1}$ with $abm_{1}\in N_{1},$ so $(a,1)(b,1)(m_{1
	,0)=(abm_{1},0)\in N.$ Hence $(s,1)(a,1)(b,1)\in(N_{1}\times M_{2}:M_{1}\times
	M_{2})$ or $(s,1)(m_{1},0)\in M$-$rad(N_{1}\times M_{2}).$ If
	$(s,1)(a,1)(b,1)\in(N_{1}\times M_{2}:M_{1}\times M_{2}),$ it is clear that
	$sab\in(N_{1}:M_{1}).$ Now suppose $(s,1)(m_{1},0)\in M$-$rad(N_{1}\times
	M_{2}).$ By \cite[Lemma 2.10]{Ebrahimi}, we have $(s,1)(m_{1},0)\in
	M_{1}$-$rad(N_{1})\times M_{2},$ and thus $sm_{1}\in M_{1}$-$rad(N_{1}).$ We conclude that $N_{1}$ is an $S$-1-absorbing primary submodule. Now assume that $N_{1}$ is an $S$-1-absorbing primary submodule of $M_{1}$ and let $s\in S$ be an $S$-element of $N_{1}.$ Let $(a,a^{\prime}), (b,b^{\prime})$ be non-units of $R\times R$
	and $(m,m^{\prime})\in M$ with $(a,a^{\prime})(b,b^{\prime})(m,m^{\prime})\in
	N.$ Then $abm\in N_{1}$ and so either $sab\in(N_{1}:M_{1})$ or $sm_{1}\in
	M_{1}$-$rad(N).$ If $sab\in N_{1},$ we conclude that $(s,1)(a,a^{\prime
	})(b,b^{\prime})=(sab,a^{\prime}b^{\prime})\in(N_{1}\times M_{2}:M_{1}\times
	M_{2}),$ now if $sab\in M_{1}$-$rad(N_{1}),$ we get $(s,1)(a,a^{\prime
	})(m,m^{\prime})=(sam,sa^{\prime}m^{\prime})\in M_{1}$-$rad(N_{1})\times
	M_{2}=M$-$rad(N_{1}\times M_{2})$ by \cite[Lemma 2.10]{Ebrahimi}. Hence, $N$ is an
	$S\times\{1\}$-1-absorbing primary submodule of $M.$

	(2) The similar proof of (1).

\end{proof}

\begin{lemma}
	\label{mrad}  Let $N $ be a submodule of $M $ and $S $ be a multiplicatively
	closed subset of $R. $ If $(P:M) \cap S= \emptyset$ for all prime submodule $P
	$ containing $N, $ then $(M$-$rad(N):s)=M$-$rad(N)=M$-$rad(N:s) $ for any $s\in S. $
\end{lemma}

\begin{proof}
	The inclusions $M$-$rad(N)\subseteq M$-$rad(N:s) $ and $M$-$rad(N)\subseteq
	(M$-$rad(N):s) $ are clear. Let $m\in(N:s) $ and suppose that $P $ is a prime
	submodule of $M $ containing $N. $ Then $sm\in N\subseteq P $ implies directly
	$m\in P $ due to our assumption. Thus, $(N:s)\subseteq P $ and we conclude
	$M$-$rad(N)=M$-$rad(N:s). $ Now, let $m\in(M$-$rad(N):s). $ Then $sm\in M$-$rad(N) $
	which means that $sm\in P $ for all prime submodules $P $ containing $N. $
	Again we have $m\in P $ by our assumption. It follows that $m\in M$-$rad(N) $
	and consequently we get $(M$-$rad(N):s)\subseteq M$-$rad(N). $
\end{proof}

In the following proposition, we denote $Z_{I}(R) $ by the set $\{x\in R \mid
xy\in I \text{ for some } y\in R\setminus I\}. $

\begin{proposition}
	\label{Ns}  Let $N $ be a submodule of $M $ and $S $ be a multiplicatively
	closed subset of $R $ provided $(P:M) \cap S= \emptyset$ for all prime
	submodule $P $ containing $N. $ If $Z_{(N:M)}(R) \cap S= \emptyset, $ then
	$(N:s) $ is a 1-absorbing primary submodule of $M $ for all $s\in S. $
\end{proposition}

\begin{proof}
	Suppose that $abm\in(N:s) $ for some non-unit $a,b\in R $ and $m\in M. $
	Hence, $sabm\in M $ and there exists $s^{\prime}\in S $ such that either
	$s^{\prime}sab\in(N:M) $ or $s^{\prime}m\in M$-$rad(N). $ If $s^{\prime
	sab\in(N:M), $ then we conclude $ab\in(N:M)\subseteq((N:s):M) $ since
	$Z_{(N:M)}(R) \cap S= \emptyset. $ Now, assume that $s^{\prime}m\in M$-$rad(N).
	$ Then we have $m\in(M$-$rad(N):s^{\prime})=M$-$rad(N)=M$-$rad(N:s) $ by Lemma
	\ref{mrad}, so we are done.
\end{proof}

\section{$S$-1-absorbing Primary Avoidance Theorem}\label{sec3}

In this section, $S $-1-absorbing primary avoidance theorem is proved.
Throughout this section, let $M $ be a finitely generated multiplication $R
$-module, $N, N_{1},N_{2}, \dots, N_{n} $ be submodules of $M $ and $S $ be a
multiplicatively closed subset of $R. $ Recall that a covering $N\subseteq
N_{1} \cup N_{2} \cup\dots\cup N_{n}$ is said to be efficient if no $N_{k} $
is unnecessary. Besides, if none of the $N_{k} $ may be excluded, the union
$N= N_{1} \cup N_{2} \cup\dots\cup N_{n}$ is efficient \cite{Lu}. We first
state some theorems that we will require.

\begin{lemma}
	(\cite{HM}, Corollary 1)\label{Cor1}  Let $R $ be a ring, $S \subseteq R $ a
	multiplicatively closed set and $P $ an ideal of $R $ disjoint with $S $. Then
	$P $ is $S $-prime if and only if there exists $s \in S $, such that for all
	$I_{1}, \dots, I_{n} $ ideals of R, if $I_{1} \dots I_{n} \subseteq P $, then
	$sI_{j} \in P $ for some $j\in\{1,\dots, n\}. $
\end{lemma}

\begin{lemma}
	(\cite{HM}, Proposition 4)\label{Prop4}  Let $R $ be a ring, $S \subseteq R $
	a multiplicatively closed set and $P $ an ideal of $R $ disjoint with $S $.
	Then $P $ is $S $-prime if and only if there exists $s \in S $ such that for
	all $x_{1}, \dots, x_{n} \in R $, if $x_{1} \dots x_{n} \in P $, then $sx_{j}
	\in P $ for some $j\in\{1,\dots, n\}. $
\end{lemma}

\begin{theorem}
	\label{A1} Let $N\subseteq N_{1}\cup N_{2}\cup\dots\cup N_{n}$ be an efficient
	covering of submodules $N_{1},N_{2},\dots,N_{n}$ of $M$ where $n>2$. Assume
	that $\sqrt{(N_{k}:m)}$ is an ideal of $R$ disjoint with $S$ {\ for all $1\leq
		k\leq n$ and all $m\in M\setminus P$ where $P$ is any prime submodule of $M$
		containing $N_{k}.$} If $\sqrt{(N_{j}:M)}\nsubseteq(\sqrt{(N_{k}:m)}:s)$ for
	all $s\in S$ whenever $j\neq k,$ then no $N_{k}$ is an $S$-1-absorbing primary
	submodule of $M.$
\end{theorem}

\begin{proof}
	It can be easily seen that $N=(N\cap N_{1})\bigcup(N\cap N_{2})\bigcup
	\dots\bigcup(N\cap N_{n})$ is an efficient union since $N\subseteq N_{1} \cup
	N_{2} \cup\dots\cup N_{n}$ is efficient.  Thus, there exists an element $m_{k}
	\in N \setminus N_{k}$ for all $k\leq n. $ From (\cite{Lu}, Lemma 2.2), it is
	known that $\bigcap_{j\neq k}(N\cap N_{j})\subseteq N \cap N_{k}. $ Now assume
	that $N_{k} $ is an $S $-1-absorbing primary submodule of $M $ for some
	$k\in\{1,\dots, n\}. $ Then, $(N_{k}:m) $ is an $S $-1-absorbing primary ideal
	of $R $ by Lemma \ref{d}. Furthermore, by Lemma \ref{lrad}, $\sqrt{(N_{k}:m)}
	$ is an $S $-prime ideal of $R. $ By our hypothesis, $s\sqrt{(N_{j}:M)}
	\nsubseteq(\sqrt{(N_{k}:m)} $ and then there exists a non-unit element
	$a_{j}\in\sqrt{(N_{j}:M)} $ such that $sa_{j}\notin\sqrt{(N_{k}:m)} $ for all
	$s\in S $ and all $m\in M \setminus N_{k}. $ (If $a_{j} $ is unit, then
	$\sqrt{(N_{j}:M)}=R $ and hence $N_{j}=M $ which contradicts with our
	assumption that the covering is efficient.) So, there exists a positive
	integer $n_{j} $ such that $a_{j}^{n_{j}}\in(N_{j}:M) $ where $j\neq k. $ From
	Lemma \ref{Cor1}, we also have $\prod_{j\neq k} \sqrt{(N_{j}:M)}
	\nsubseteq\sqrt{(N_{k}:m)} $ due to $\sqrt{(N_{k}:m)} $ is $S $-prime. Now let
	us say $a= \prod_{j\neq k}a_{j}$ and $t=max\{n_{j}\}_{j\neq k}. $ Then we get
	$a^{t} \in(N_{j}:M) $ and it follows that $a^{t}m_{k} \in N \cap N_{j} $ for
	every $j\neq k. $ However, $a \notin\sqrt{(N_{k}:m)} $ by Lemma \ref{Prop4}
	and this implies that $a^{t}m\notin N_{k} $ for all $m\in M \setminus N_{k}. $
	As a result, $a^{t}m_{k} \notin N_{k}$ because $m_{k} \notin N_{k}$ as well.
	This shows that $a^{t}m_{k} \in\bigcap_{j\neq k}(N\cap N_{j})\setminus N \cap
	N_{k}, $ which is a contradiction.
\end{proof}

\begin{theorem}
	[$S $-1-absorbing primary avoidance theorem] Let $N, N_{1},N_{2}, \dots, N_{n}
	$ be submodules of $M $ such that $N\subseteq N_{1} \cup N_{2} \cup\dots\cup
	N_{n}$ where at most two of $N_{1},N_{2}, \dots, N_{n} $ are not $S
	$-1-absorbing primary submodules. Suppose that $\sqrt{(N_{k}:m)} \cap
	S=\emptyset$ for all $m\in M \setminus N_{k} $ where $1\leq k\leq n $. If
	$\sqrt{(N_{j}:M)} \nsubseteq(\sqrt{(N_{k}:m)}:s) $ for all $s\in S $ whenever
	$j\neq k, $ then $N \subseteq N_{k} $ for some $k. $
\end{theorem}

\begin{proof}
	We can assume that the covering $N\subseteq N_{1} \cup N_{2} \cup\dots\cup
	N_{n}$ is efficient because any cover including submodules of $M $ can be
	reduced to an efficient covering by eliminating any superfluous terms and
	therefore $n \neq2 $. Since there is no $S $-1-absorbing primary submodule
	$N_{k} $ from Theorem \ref{A1}, $n<2 $. Hence we obtain $N \subseteq N_{k} $
	for some $k \in\{1,\dots, n\}. $
\end{proof}

\begin{corollary}
	Let $N $ be a submodule of $M. $ If the $S $-1-absorbing primary avoidance
	theorem holds for $M $, then the theorem holds also for $M/N. $
\end{corollary}

\begin{proof}
	Let $K/N, N_{1}/N, N_{2}/N, ..., N_{n}/N $ be submodules of $M/N $ such that
	at most two of $N_{1}/N, N_{2}/N, ..., N_{n}/N $ are not $S $-1-absorbing
	primary and $K/N \subseteq N_{1}/N \cap N_{2}/N \cap\dots\cap N_{n}/N $ where
	$n\geq2 $ . Thus, $K \subseteq N_{1} \cap N_{2} \cap\dots\cap N_{n} $ and at
	most two of $N_{1}, N_{2}, ..., N_{n} $ are not $S $-1-absorbing primary by
	Corollary 1. Assume that $\sqrt{N_{j}/N:M/N}\nsubseteq(\sqrt{N_{k}/N:m+N}:s)$
	for all $s\in S $ and all $m+N \in M/N\setminus N_{k}/N $ whenever $j\neq k. $
	Then $\sqrt{(N_{j}:M)} \nsubseteq(\sqrt{(N_{k}:m)}:s) $ for all $s\in S $.
	Since the $S $-1-absorbing primary avoidance theorem is satisfied by the
	submodules of $M, $ the rest of the proof is immediate.
\end{proof}

\section{Idealization of $S$-1-absorbing primary submodules}\label{sec4}

Let $M$ be a unitary $R$-module. Recall from \cite{Nagata} the idealization
$R(+)M$ of $M$ is the direct sum $R(+)M=R\oplus M$ which is a commutative ring
with componentwise addition and multiplication $(r_{1},m_{1})(r_{2},m_{2
)=(r_{1}r_{2},r_{1}m_{2}+r_{2}m_{1}).$  It is well known that if $N$ is a
submodule of $M$ with $IM\subset N,$ then $I(+)N$ is an ideal of $R(+)M.$

\begin{proposition}
	\label{id}Let $S$ be a multiplicatively closed subset of $R$, $I$ be an ideal
	of $R$ with $ I\cap S=\emptyset.$ Then the following statements are equivalent:

	\begin{enumerate}
		\item $I$ is an $S$-1-absorbing primary ideal of $R$.

		\item $I(+)M$ is an $S(+)0$-1-absorbing primary ideal of $R(+)M$.

		\item $I(+)M$ is an $S(+)M$-1-absorbing primary ideal of $R(+)M.$
	\end{enumerate}
\end{proposition}

\begin{proof}
	(1)$\Rightarrow$ (2) It is clear that $(I(+)N) \cap(S(+)0)=\emptyset$ if and
	only if $S\cap I=\emptyset.$ Suppose that $I$ is an $S$-1-absorbing primary ideal
	of $R$ and $s$ be an $S$-element of $I.$ Let $(a,m)(b,n)(c,p)\in I(+)M$ where
	$(a,m),(b,n),(c,p)$ are non-unit elements of $R(+)M.$ Since $U(R(+)M)=U(R)(+)M,$ we
	conclude that $a,b,c$ are non-units in $R$ and $abc\in I.$ Then either $sab\in
	I$ or $sc\in\sqrt{I}.$ If $sab\in I,$ then $(s,0)(a,m)(b,n)=(sab,san+sbm)\in
	I(+)M.$ If $sc\in\sqrt{I},$ then $(s,0)(c,p)=(sc,0)\in\sqrt{I(+)M}$ due to $\sqrt{I(+)M}=\sqrt{I}(+)M$. Thus $I(+)M$ is an $S(+)0$-1-absorbing
	primary ideal of $R(+)M.$

	(2)$\Rightarrow$ (3) Is clear since $S(+)0\subset S(+)M.$

	(3)$\Rightarrow$ (1) Assume that $I(+)M$ is an $S(+)M$-1-absorbing primary
	ideal of $R(+)M$. Let $(s,m)$ be an $S(+)M$-element of $I(+)M$ and let $abc\in
	I$ for non-unit elements $ a,b,c $ of $R.$ Then $(a,0),(b,0),(c,0)$ are also non-unit elements of $R(+)M$ since $U(R(+)M)=U(R)(+)M$. Besides, we have $(a,0)(b,0)(c,0)\in I(+)M.$ So, either $(s,m)(a,0)(b,0)=(sab,abm)\in I(+)M$ or $(s,m)(c,0)\in \sqrt{I(+)M}$ which yields $sab\in I$ or $sc\sqrt{I}.$ Hence $I$ is an $S(+)M$-1-absorbing primary ideal of $R.$
\end{proof}

\begin{proposition}
	\label{id}Let $S$ be a multiplicatively closed subset of $R$ and $I$ be an ideal of $R$ with $I\cap S=\emptyset.$ Suppose $N$ is a proper submodule of $M$ such that $IM\subset N.$ Then $(1)\Longrightarrow (2)\Longrightarrow(3)$ is satisfied for the below statements:

	\begin{enumerate}
		\item $I(+)N$ is an $S(+)0$-1-absorbing primary ideal of $R(+)M$.

		\item $I(+)N$ is an $S(+)M$-1-absorbing primary ideal of $R(+)M.$

		\item $I$ is an $S$-1-absorbing primary ideal of $R$.
	\end{enumerate}
\end{proposition}

\begin{proof}
	$(1)\Longrightarrow(2)$ It is clear since $S(+)0\subset S(+)M.$

	$(2)\Longrightarrow(3)$ Let $abc\in I$ for non-unit elements $a,b,c\in R.$ We
	have $(a,0)(b,0)(c,0)\in I(+)N$ which is $S(+)M$-1-absorbing primary ideal of
	$R(+)M$ where $(a,0),(b,0),(c,0)$ are non-units in $R(+)M.$ Then either
	\[
	(s,m)(a,0)(b,0)=(sab,abm)\in I(+)N
	\]
	o

	\[
	(s,m)(c,0)=(sv,cm)\in\sqrt{I(+)N}=\sqrt{I}(+)M.
	\]
	where $(s,m)$ is an $S(+)M$-element of $I(+)N.$ Then either $sab\in I$ or
	$sc\in\sqrt{I}.$ Hence $I$ is an $S$-1-absorbing primary ideal of $R.$
\end{proof}

\begin{example}
	The implication $(3)\Longrightarrow(1)$ is not true in general case. If $I$ is an
	$S$-1-absorbing primary ideal, then $I(+)N$ (where $N$ a proper submodule of $M$)
	is not necessarily $S(+)0$-1-absorbing primary ideal of $R(+)M.$ Take
	$R=\mathbf{Z}=M$ and $S=\{\mp1\}.$ Then $I=\{0\}$ is an $S$-1-absorbing
	primary ideal of $\mathbf{Z},$ since $(0(+)6\mathbf{Z}:(0,1))=6\mathbf{Z}(+)\mathbf{Z}$
	is not an $S(+)0$-prime ideal. We conclude that
	$0(+)6\mathbf{Z}$ is not an $S(+)0$-1-absorbing primary ideal of $\mathbf{Z
	(+)\mathbf{Z}$ by Lemma \ref{lrad}.
\end{example}

\section{Amalgamation of $S$-1-absorbing primary submodules}\label{sec5}

Let $A$ and $B$ be commutative rings with unity, let $J$ be an ideal of $B$
and let $f:A\rightarrow B$ be a ring homomorphism. In this setting, we can
consider the following subring of $A\times B:$
\[
A\bowtie^{f}J:=\left\{ (a,f(a)+j)\mid a\in A, j\in J\right\}
\]
called the \textit{amalgamation of $A$ with $B$ along $J$ with respect to $f
}. This construction is a generalization of the amalgamated duplication of a
ring along an ideal (cf., for instance \cite{PP, AMES, BMT, MMSN, DF, MCM}).

It is clear that a subset $S$ of $A$ is a multiplicatively closed subset of
$A$ if and only if $S\bowtie^{f}J:=\{(s,f(s)+j)\mid s\in S;j\in J\}$ is a
multiplicatively closed subset of $A\bowtie^{f}J.$

\begin{lemma}
	\label{amal1}

	If $I$ is an ideal of $A,$ then
	\[
	\sqrt{I\bowtie^{f}J}=\sqrt{I}\bowtie^{f}J.
	\]

\end{lemma}

\begin{proof}

	Let $(a,f(a)+j)\in A\bowtie^{f}J$ such that $(a,f(a)+j)\in\sqrt{I\bowtie^{f
		J}.$ Then there exists a nonzero integer $n$ such that $(a,f(a)+j)^{n
	=(a^{n},f(a^{n})+j^{\prime})\in I\bowtie^{f}J$ where $j^{\prime
	=\displaystyle \sum_{k=0}^{n-1}\left(
	\begin{array}
	[c]{c
	n\\
	k\\
	\end{array}
	\right) f(a^{k})j^{n-k}\in J.$ Then $a^{n}\in I$, so $a\in\sqrt{I}$ and
	$(a,f(a)+j)\in\sqrt{I}\bowtie^{f}J$. Hence, we conclude that $\sqrt{I\bowtie^{f}J
	\subset\sqrt{I}\bowtie^{f}J.$ Conversely, let $(a,f(a)+j)\in\sqrt{I
	\bowtie^{f}J$. Then there exists $n\in\mathbb{N}^{*}$ such that $a^{n}\in I.$
	On the other hand $(a,f(a)+j)^{n}=(a^{n},f(a^{n})+j^{\prime }).$ Then
	$(a,f(a)+j)\in\sqrt{I\bowtie^{f}J}$, which yields $\sqrt{I}\bowtie
	^{f}J\subset\sqrt{I\bowtie^{f}J}.$ Finally, we conclude $\sqrt{I\bowtie^{f}J}=\sqrt
	{I}\bowtie^{f}J.$
\end{proof}

\begin{proposition}
Let $ S $ be a multiplicatively closed subset and $ I, J $ be ideals of $ A $. The following statements are equivalent:

	\begin{enumerate}
		\item $I$ is an $S$-1-absorbing primary ideal of $A.$

		\item $I\bowtie^{f}J$ is an $S\bowtie^{f}J$-1-absorbing ideal of $A\bowtie
		^{f}J.$
	\end{enumerate}
\end{proposition}

\begin{proof}

	Assume that $I\bowtie^{f}J$ is an $S\bowtie^{f}J$-1-absorbing primary ideal of
	$A\bowtie^{f}J$ and let $(s,f(s)+j)$ be an $S\bowtie^{f}J$-element of
	$I\bowtie^{f}J.$ Take $abc\in I$ for non-unit elements $a,b,c$ of $A.$ Then
	$(a,f(a)),(b,f(b)),(c,f(c))$ are also non-unit elements of $A\bowtie^{f}J$ and
	$(a,f(a))(b,f(b))(c,f(c))\in I\bowtie^{f}J.$ Since $I\bowtie^{f}J$ is an
	$S\bowtie^{f}J$-1-absorbing primary submodule, we get either
	$(s,f(s)+j)(a,f(a))(b,f(b))\in I\bowtie^{f}J$ or $(s,f(s)+j)(c,f(c))\in
	\sqrt{I\bowtie^{f}J}=\sqrt{I}\bowtie^{f}J$ by Lemma \ref{amal1}. Then
	either $sab\in I$ or $sc\in\sqrt{I}.$  Thus $I$ is an $S$-1-absorbing primary
	ideal of $A.$

	Conversely, suppose that $I$ is an $S$-1-absorbing primary ideal of $A$ and let
	$s\in S$ be an $S$-element of $I.$ Let $(a,f(a)+j),(b,f(b)+k),(c,f(c)+l)$ be
	non-unit elements of $A\bowtie^{f}J$ such that
	$(a,f(a)+j)(b,f(b)+k)(c,f(c)+l)\in I\bowtie^{f}J.$ Then $abc\in I.$It is necessary that one of the $a,b,c$ is non-unit, otherwise, $abc$ is unit and $abc\in I,$ a contradiction. Since $I$ is an $S$-1-absorbing primary submodule of $A,$ we get either
	$sab\in I$ or $sc\in\sqrt{I}$ which gives either $(s,f(s))(a,f(a)+j)(b,f(b)+k)\in
	I\bowtie^{f}J$ or $(s,f(s))(c,f(c)+l)\in\sqrt{I}\bowtie^{f}J=\sqrt
	{I\bowtie^{f}J}$ by Lemma \ref{amal1}. Hence $I\bowtie^{f}J$ is an
	$S\bowtie^{f}J$-1-absorbing primary ideal of $A\bowtie^{f}J.$
\end{proof}

\begin{corollary}
	Let $R$ be a ring and $I, K$ be ideals of $R$ with $K\cap S=\emptyset.$ Then
	the following statements are equivalent:

	\begin{enumerate}
		\item $K$ is an $S$-1-absorbing primary ideal of $R.$

		\item $K\bowtie I$ is an $S\bowtie I$-1-absorbing primary ideal of $R\bowtie I.$
	\end{enumerate}
\end{corollary}

\begin{proof}
	It is easy to see that $(K\bowtie I)\cap(S\bowtie I)=\emptyset.$ The rest of
	the proof follows from the previous proposition.
\end{proof}

Let $S_{2}$ be a multiplicatively closed subset of the ring $f(A)+J.$ Consider the set
\[
\overline{S_{2}}^{f}:=\{(r,f(r)+j)\mid j\in J, r\in A, f(r)+j\in S_{2}\}.
\]
It is clear that $\overline{S_{2}}^{f}$ is a multiplicatively closed subset of
$A\bowtie^{f}J.$

If $K$ is an ideal of $f(A)+J$, then the set
\[
\overline{K}^{f}:=\{(a,f(a)+j)\mid a\in A,j\in J,f(a)+j\in K\}
\]
is an ideal of $A\bowtie^{f}J.$

\begin{lemma}
	\label{amal2}

	Under the above notations, we have
	\[
	\sqrt{\overline{K}^{f}}=\overline{\sqrt{K}}^{f}.
	\]

\end{lemma}

\begin{proposition}
	Let $f:A\longrightarrow B$ be a ring homomorphism, $J, K$ be ideals of $B$ and
	$S_{2}$ be a multiplicatively closed subset of $f(A)+J$ with $K\cap
	S_{2}=\emptyset.$ Let
	\[
	\overline{K}^{f}=\{(a,f(a)+j)\mid a\in A, j\in J, f(a)+j\in K\}.
	\]
	Then the following statements are equivalent:

	\begin{enumerate}
		\item $K$ is an $S_{2}$-1-absorbing primary ideal of $f(A)+J.$

		\item $\overline{K}^{f}$ is an $\overline{S}^{f}_{2}$-1-absorbing primary ideal
		of $A\bowtie^{f}J.$
	\end{enumerate}
\end{proposition}

\begin{proof}

	Assume that $\overline{K}^{f}$ is an $\overline{S}^{f}$-1-absorbing primary
	ideal of $A\bowtie^{f}J$ and let $(s,f(s)+p)$ be an $\overline{S}^{f}$-element of
	$\overline{K}^{f}.$ Let $(f(a)+j)(f(b)+k)(f(c)+l)\in K$ for non-unit elements
	$f(a)+j,f(b)+k,f(c)+l\in f(A)+J.$ Then $(a,f(a)+j),(b,f(b)+k),(c,f(c)+l)$ are
	non-unit elements of $A\bowtie^{f}J$ and $(a,f(a)+j)(b,f(b)+k)(c,f(c)+l)
	\in\overline{K}^{f}.$ Thus, we have either $(s,f(s)+p)(a,f(a)+j)(b,f(b)+k)\in
	\overline{K}^{f}$ or $(s,f(s)+p)(c,f(c)+l)\in\sqrt{\overline{K}^{f}}.$ In the
	first case, we have $(f(s)+p)(f(a)+j)(f(b)+k)\in K.$ In the second case, we have
	$(f(s)+p)(f(c)+k)\in\sqrt{\overline{K}^{f}}=\overline{\sqrt{K}}^{f}$ by Lemma \ref{amal2} and hence
	$(f(s)+p)(f(c)+k)\in\sqrt{K}.$ We conclude that $K$ is an $S_{2}$-1-absorbing
	primary ideal of $B.$ Conversely, suppose that $K$ is an $S_{2}$-1-absorbing
	primary ideal of $B$ and let $f(s)+p$ an $S_{2}$-element of $K.$ Let
	$(a,f(a)+j)(b,f(b)+k)(c,f(c)+l)\in\overline{K}^{f}$ for non-unit elements
	$(a,f(a)+j),(b,f(b)+k),(c,f(c)+l)\in A\bowtie^{f}J.$ Then we get $(abc,
	(f(a)+j)(f(b)+k)(f(c)+l))\in\overline{K}^{f}.$ Now suppose that $f(a)+j, f(b)+k,
	f(c)+l$ are non-unit elements of $f(A)+J.$ Since $K$ is an $S_{2}$-1-absorbing primary
	ideal of $f(A)+J,$ we conclude that either $(f(s)+p)(f(a)+j)(f(b)+k)\in K$ or
	$(f(s)+p)(f(c)+l)\in\sqrt{K}.$ In the first case, we have
	$(s,f(s)+p)(a,f(a)+j)(b,f(b)+k)\in\overline{K}^{f}.$ In the second case, we
	have $(s,f(s)+p)(c,f(c)+l)\in\sqrt{\overline{K}^{f}}.$ Now, if one of $f(a)+j, f(b)+k, f(c)+l$ is unit, then the claim is clear. Hence
	$\overline{K}^{f}$ is an $\overline{S_{2}}^{f}$-1-absorbing primary ideal of
	$A\bowtie^{f}J.$
\end{proof}

An $R$-module can be regarded as an $R\bowtie I$-module via the natural projection $R\bowtie I\longrightarrow R; (r,r+i)\longmapsto r,$ and an $R\bowtie I$-module can also be regarded as an $R$-module via the natural injection $\iota:R\longrightarrow
R\bowtie I; r\longmapsto(r,r).$

Let $R$ be a ring, $I$ an ideal of $R$, $M$ an $R$-module and set
\[
M\bowtie I:=\{(m,m^{\prime})\in M\times M\mid m-m^{\prime}\in IM\},
\]
which is an $R\bowtie I$-module with the multiplication given by
$$ (r,r+i)(m,m^{\prime})=(rm,(r+i)m^{\prime}) $$
where $ r\in R $, $ i\in I $ and $ (m,m^{\prime})\in M\bowtie I. $
Then the set $M\bowtie I$ is called the \textit{duplication of the $R$-module $M$ along the
	ideal $I.$}

Let $N$ be a submodule of $M.$ It is clear that
\[
N\bowtie I:\{(n,m)\in N\times M\mid n-m\in IM\}
\]
and
\[
\overline{N}:=\{(m,n)\in M\times N\mid n-m\in IM\}
\]
are submodules of $M\bowtie I.$ If $S$ is a multiplicatively closed subset of
$R,$ then it is obvious that the sets $S\bowtie I:=\{(s,s+i)\mid s\in S,i\in I\}$ and
$\overline{S}:=\{(s,s+i)\mid i\in I,s+i\in S\}$ are multiplicatively closed
subsets of $R\bowtie I.$
\begin{lemma}
	\label{lduplication1}

	Let $N$ be a submodule of $M.$ Then
	\[
	(N\bowtie I:_{R\bowtie I}M\bowtie I)=(N:_{R}M)\bowtie I.
	\]

\end{lemma}

\begin{proof}
	Let $(a,a+i)\in(N\bowtie I:_{R\bowtie I}M\bowtie I)$. Then $(a,a+i)(M\bowtie I)\subset N\bowtie I.$ Clearly,
	$aM\subset N$ and thus $(a,a+i)\in(N:_{R}M)\bowtie I.$ Conversely, suppose
	that $(a,a+i)\in(N:_{R}M)\bowtie I$ and take $(m,m^{\prime})\in M\bowtie I.$
	We have $(a,a+i)(m,m^{\prime})=(am,(a+i)m^{\prime})\in M\times M$ and
	$a(m-m^{\prime})-im^{\prime}\in IM$ since $m-m^{\prime}\in IM.$ Hence
	$(a,a+i)\in (N\bowtie I:_{R\bowtie I}M\bowtie I).$ Thus, we conclude that $(N\bowtie
	I:_{R\bowtie I}M\bowtie I)=(N:_{R}M)\bowtie I.$
\end{proof}

\begin{lemma}
	\label{lduplication2}

	Let $L$ be a submodule of $M.$ Then the following statements are equivalent:

	\begin{enumerate}
		\item $L$ is a prime submodule of $M.$

		\item $L\bowtie I$ is a prime submodule of $M\bowtie I.$
	\end{enumerate}
\end{lemma}

\begin{proof}
	Assume that $L\bowtie I$ is a prime submodule of $M\bowtie I.$ Let $am\in L$
	for $a\in R$ and $m\in M.$ Then $(a,a)(m,m)\in L\bowtie I.$ Since $L\bowtie I$
	is a prime submodule of $M\bowtie I,$ we get either $(a,a)\in(L\bowtie
	I:_{R\bowtie I}M\bowtie I)$ which implies $a\in (N:M)$ by Lemma
	\ref{lduplication1}, or $(m,m)\in L\bowtie I$ which follows $m\in L.$ Hence $L$
	is a prime submodule of $M.$ Conversely, assume that $L$ is a prime submodule of $ M $
	and let $(a,a+i)(m,m^{\prime})\in L\bowtie I$ for $(a,a+i)\in R\bowtie I$ and
	$(m,m^{\prime})\in M\bowtie I.$ Thus $(am,(a+i)m^{\prime})\in L\bowtie I$ and we
	conclude that $am\in L.$ Due to the fact that L is a prime submodule, we obtain
	either $a\in(N:M)$ or $m\in L.$ In the first case, we have $(a,a+i)\in
	(N:M)\bowtie I=(N\bowtie I:_{R\bowtie I}M\bowtie I).$ In the second case, we
	have $(m,m^{\prime})\in L\bowtie I,$ since $m-m^{\prime}\in IM.$ Thus,
	$L\bowtie I$ is a prime submodule of $M\bowtie I.$
\end{proof}

\begin{lemma}
	\label{lduplication3} Let $N$ be a submodule of $M.$ Then the prime submodule
	which contains $N\bowtie I$ are exactly of the form $L\bowtie I$ where $L$ is
	a prime submodule containing $N.$ Moreover, $ M\bowtie I$-$rad(N\bowtie I)=M$-$rad(N)\bowtie I. $

\end{lemma}

\begin{proof}
	Let $\varphi:M\bowtie I\longrightarrow M;(m,m^{\prime})\longmapsto m$ be an
	$R\bowtie I$-epimorphism with $\ker\varphi=0\times IM.$ Then $M\bowtie
	I/0\times IM\simeq M.$ So, if $L$ is a submodule of $M\bowtie I$ such that
	$0\times IM\subset I,$ then there exists a submodule $K$ of $M$ such that $L/0\times
	IM\simeq K,$ which implies $L=K\bowtie I.$ From Lemma \ref{lduplication2}, we
	conclude that $L$ is a prime submodule of $M\bowtie I$ if and only if $K$ is a
	prime submodule of $M.$ Now, if $L$ is a prime submodule of $M\bowtie I$ such
	that $N\bowtie I\subset L,$ then $0\times IM\subset L,$ and so there exits a
	prime submodule $K$ of $M$ with $L=K\bowtie I.$

	Now, it is clear that $\bigcap_i(K_i\bowtie
	I)=\left( \bigcap_iK_i\right) \bowtie I$ where $ K_i $'s are the submodules of $ M $ and
	consequently we have $M\bowtie I$-$rad(N\bowtie I)=M$-$rad(N)\bowtie I.$
\end{proof}

\begin{theorem}
	Let $N$ be a submodule of $M$ with $(N:_{R}M)\cap S=\emptyset.$ Then the
	following statements are equivalent:

	\begin{enumerate}
		\item $N$ is an $S$-1-absorbing primary submodule of $M.$

		\item $N\bowtie I$ is an $S\bowtie I$-1-absorbing primary submodule of $M\bowtie
		I.$
	\end{enumerate}
\end{theorem}

\begin{proof}
	From Lemma \ref{lduplication1}, we clearly note that $(N\bowtie
	I:_{R\bowtie I}M\bowtie I)\cap S\bowtie I=\emptyset$ if and only if
	$(N:_{R}M)\cap S=\emptyset.$ Now suppose that $N\bowtie I$ is an $S\bowtie I$-1-absorbing
	primary submodule of $M\bowtie I$. Let $(s,s+i)$ be an $S\bowtie I$-element of
	$N\bowtie I$ and let $abm\in N$ for non-unit elements $a,b\in R$ and $m\in M.$
	Then $(a,a)(b,b)(m,m)\in N\bowtie I.$ Since $(a,a),(b,b)$ are non-units and
	$N\bowtie I$ is an $S\bowtie I$-1-absorbing primary submodule of $M\bowtie I,$ we
	have either
	\[
	(s,s+i)(a,a)(b,b)\in(N\bowtie I:M\bowtie I)
	\]
	o

	\[
	(s,s+i)(m,m)\in M\bowtie I-\text{rad}(N\bowtie I).
	\]
	In the first case, we have $sab\in (N:M)$ by Lemma \ref{lduplication1}. In the second case, $M\bowtie I$-$rad(N\bowtie I)=M$-$rad(N)\bowtie I$ by Lemma \ref{lduplication3} and we conclude that  $(sm,(s+i)m)\in M$-$rad(N)\bowtie I,$  and so $sm\in M$-$rad(N).$ If such an $ N $ does not exists, it is well known that $M$-$rad(N)=M.$ Then $N$ is an $S$-1-absorbing primary submodule of $M.$
	Conversely, assume that $N$ is an $S$-absorbing primary submodule of $M$. Let $s\in S$ be an $S$-element of $N$ and let $(a,a+i)(b,b+i)(m,m')\in N\bowtie I$ for non-unit elements $(a,a+i),(b,b+i)\in R\bowtie I$ and $(m,m')\in M\bowtie I.$ Then $abm\in N.$ If $a$ or $b$ is unit, the claim is clear. Now we can assume $a$ and $b$ are non-unit elements of $R.$ Due to the fact that $N$ is an $S$-1-absorbing primary submodule of $R,$ we get either $sab\in (N:M)$ or $sm\in M$-$rad(N).$ In the first case, we conclude that $(s,s)(a,a+i)(b,b+i)\in(N:M)\bowtie I=(N\bowtie I:M\bowtie I).$ In the second case, we have $(sm,sm'+im')\in M$-$rad(N)\bowtie I,$ since $r(m-m')-im'\in IM.$ Thus, the claim follows by $M$-$rad(N)\bowtie I=M\bowtie I$-$rad(N\bowtie I).$

\end{proof}

redpajama_set_name: RedPajamaArXiv